Question 14 Marks
Evaluate the following limits in Exercise:$\lim\limits_{\text{x}\rightarrow-2}\frac{\frac{1}{\text{x}}+\frac{1}{2}}{\text{x}+2}$
Answer$\lim\limits_{\text{x}\rightarrow-2}\frac{\frac{1}{\text{x}}+\frac{1}{2}}{\text{x}+2}$At x = –2, the value of the given function takes the form ${\frac{0}{0}}$.
$\text{Now},$$\lim\limits_{\text{x}\rightarrow-2}\frac{\frac{1}{\text{x}}+\frac{1}{2}}{\text{x}+2}$$=\lim\limits_{\text{x}\rightarrow-2}\frac{\Big(\frac{2+\text{x}}{2\text{x}}\Big)}{\text{x}+2}$
$\lim\limits_{\text{x}\rightarrow-2}\frac{1}{2\text{x}}$
$\frac{1}{2(-2)}$
$=\frac{-1}{4}$
View full question & answer→Question 24 Marks
Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\frac{\cos\text{x}}{1+\sin\text{x}}$
AnswerLet $\text{f(x)}=\frac{\cos\text{x}}{1+\sin\text{x}}$ By quotient rule, $\text{f}'\text{(x)}=\frac{(1+\sin\text{x})\frac{\text{d}}{\text{dx}}(\cos\text{x})-(\cos\text{x})\frac{\text{d}}{\text{dx}}(1+\sin\text{x})}{(1+\sin\text{x})^2}$ $=\frac{(1+\sin\text{x})(-\sin\text{x})-(\cos\text{x})(\cos\text{x})}{(1+\sin\text{x})^2}$ $=\frac{-\sin\text{x}-\sin^2\text{x}-\cos^2\text{x}}{(1+\sin\text{x})^2}$ $=\frac{-\sin\text{x}-(\sin^2\text{x}-\cos^2\text{x})}{(1+\sin\text{x})^2}$ $=\frac{-\sin\text{x}-1}{(1+\sin\text{x})^2}$ $=\frac{-(1+\sin\text{x})}{(1+\sin\text{x})^2}$ $=\frac{-1}{(1+\sin\text{x})}$
View full question & answer→Question 34 Marks
Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\frac{\text{x}}{\sin^{\text{n}}\text{x}}$
AnswerLet $\text{f(x)}=\frac{\text{x}}{\sin^{\text{n}}\text{x}}$By quotient rule,
$\text{f}'\text{(x)}=\frac{\sin^{\text{n}}\text{x}\frac{\text{d}}{\text{dx}}\text{x}-\text{x}\frac{\text{d}}{\text{dx}}\sin^{\text{n}}\text{x}}{\sin^{\text{2n}}\text{x}}$
It can be easily shown that $\frac{\text{d}}{\text{dx}}\sin^{\text{n}}\text{x}=\text{n}\sin^{\text{n}-1}\text{x}\cos\text{x}$
Therefore,
$\text{f}'\text{(x)}=\frac{\sin^{\text{n}}\text{x}\frac{\text{d}}{\text{dx}}\text{x}-\text{x}\frac{\text{d}}{\text{dx}}\sin^{\text{n}}\text{x}}{\sin^{\text{2n}}\text{x}}$
$=\frac{\sin^{\text{n}}\text{x}.1-\text{x}(\text{n}\sin^{\text{n}-1}\text{x}\cos\text{x})}{\sin^{\text{2n}}\text{x}}$
$=\frac{\sin^{\text{n}-1}\text{x}(\sin\text{x}-\text{nx}\cos\text{x})}{\sin^{\text{2n}}\text{x}}$
$=\frac{\sin\text{x}-\text{nx}\cos\text{x}}{\sin^{\text{n+1}}\text{x}}$
View full question & answer→Question 44 Marks
Evaluate the following limits in Exercise:$\lim\limits_{\text{x}\rightarrow0}\frac{\sin{\text{ax}}}{\sin\text{bx}},\text{a},\text{b}\neq0$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin{\text{ax}}}{\sin\text{bx}},\text{a},\text{b}\neq0$ At x = 0, the value of the given function takes the form ${\frac{0}{0}}$. $\text{Now},$$\lim\limits_{\text{x}\rightarrow0}\frac{\sin{\text{ax}}}{\sin\text{bx}}=\lim\limits_{\text{x}\rightarrow0}\frac{\Big(\frac{\sin\text{ax}}{\text{ax}}\Big)\times\text{ax}}{\Big(\frac{\sin\text{bx}}{\text{bx}}\Big)\times\text{bx}}$$=\big(\frac{\text{a}}{\text{b}}\big)\times\frac{\lim\limits_{\text{ax}\rightarrow0}\Big(\frac{\sin\text{ax}}{\text{ax}}\Big)}{\lim\limits_{\text{bx}\rightarrow0}\Big(\frac{\sin\text{bx}}{\text{bx}}\Big)}$ $\begin{bmatrix}\text{x}\rightarrow0\Rightarrow\text{ax}\rightarrow0\\ \text{and x}\rightarrow0\Rightarrow\text{bx}\rightarrow0\end{bmatrix}$
$=\big(\frac{\text{a}}{\text{b}}\big)\times\frac{1}{1}$ $\bigg[\lim\limits_{\text{y}\rightarrow0}\frac{\sin\text{y}}{\text{y}}=1\bigg]$
$=\frac{\text{a}}{\text{b}}$
View full question & answer→Question 54 Marks
Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $(\text{px+q})\Big(\frac{\text{r}}{\text{x}}+\text{s}\Big)$
AnswerLet $\text{f(x)}=(\text{px+q})\Big(\frac{\text{r}}{\text{x}}-\text{s}\Big)$ By Leibnitz product rule, $\text{f}'(\text{x})=(\text{px+q})\Big(\frac{\text{r}}{\text{x}}+\text{s}\Big)+\Big(\frac{\text{r}}{\text{x}}+\text{s}\Big)(\text{px+q)}$ $=\text{(px+q)}(\text{rx}^{-1}+\text{s})+\Big(\frac{\text{r}}{\text{x}}+\text{s}\Big)(\text{p)}$ $=(\text{px+q})(-\text{rx}^{-2})+\Big(\frac{\text{r}}{\text{x}}+\text{s}\Big)\text{p}$ $=(\text{px+q})\Big(\frac{-\text{r}}{\text{x}^2}\Big)+\Big(\frac{\text{r}}{\text{x}}+\text{s}\Big)\text{p}$ $=\frac{-\text{pr}}{\text{x}}-\frac{\text{qr}}{\text{x}^2}+\frac{\text{pr}}{\text{x}}+\text{ps}$ $=\text{ps}-\frac{\text{qr}}{\text{x}^2}$
View full question & answer→Question 64 Marks
Find the derivative of the following function from first principle. $(\text{x}-1)(\text{x}-2)$
AnswerLet $\text{f}(\text{x})=(\text{x}-1)(\text{x}-2)$. Accordingly,from the first principle, $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{x}+\text{h}-1)(\text{x}+\text{h}-2)-(\text{x-1})(\text{x}-2)}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{x}^2+\text{h}\text{x}-2\text{x}+\text{h}\text{x}+\text{h}^2-2\text{h}-\text{x}-\text{h}+2)-(\text{x}^2-2\text{x}-\text{x}+2)}{\text{h}}$$=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{h}\text{x}+\text{h}\text{x}+\text{h}^2-2\text{h}-\text{h})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(2\text{h}\text{x}+\text{h}^2-3\text{h})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}{(2\text{x+}\text{h}-3)}$$=(2\text{x}+0-3)$
$=2\text{x}-3$
View full question & answer→Question 74 Marks
Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\text{x}^4(5\sin\text{x}-3\cos\text{x})$
AnswerLet $\text{f(x)}=\text{x}^4(5\sin\text{x}-3\cos\text{x})$ By product rule, $\text{f}'\text{(x)}=\text{x}^4\frac{\text{d}}{\text{dx}}(5\sin\text{x}-3\cos\text{x})+(5\sin\text{x}-3\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{x}^4)$ $=\text{x}^4\Big[5\frac{\text{d}}{\text{dx}}(\sin\text{x})-3\frac{\text{d}}{\text{dx}}(\cos\text{x})\Big]+(5\sin\text{x}-3\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{x}^4)$ $\text{Let g(x)}=\sin\text{(x+a). Accordingly, (x+h)}=\sin(\text{x+h+a})$ $=\text{x}^4\big[5\cos\text{x}-3(-\sin\text{x})\big]+(5\sin\text{x}-3\cos\text{x})(4\text{x}^3)$ $=\text{x}^3\big[5\text{x}\cos\text{x}+3\text{x}\sin\text{x}+20\sin\text{x}-12\cos\text{x}\big]$
View full question & answer→Question 84 Marks
Find $\lim\limits_{\text{x}\rightarrow5}\text{f}(\text{x})$, where $\text{f}(\text{x})=|\text{x}|-5$
AnswerThe given function is $\text{f}(\text{x})=|\text{x}|-5$. $=\lim\limits_{\text{x}\rightarrow5^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow5^-}\big[{|\text{x}|-5}\big]$ $=\lim\limits_{\text{x}\rightarrow5}\big({\text{x}-5}\big)$ [When x>0,$|\text{x}|=\text{x}$] $=5-5$$=0$
$\lim\limits_{\text{x}\rightarrow5^+}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow5^+}\big[{|\text{x}|-5}\big]$
$=\lim\limits_{\text{x}\rightarrow5}\big[{\text{x}-5}\big]$ [When x>0,$|\text{x}|=\text{x}$] $=5-5$ $=0$ $\therefore\lim\limits_{\text{x}\rightarrow5^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow5^+}\text{f}(\text{x})=0$ Hence,$\lim\limits_{\text{x}\rightarrow5}\text{f}(\text{x})=0$
View full question & answer→Question 94 Marks
Find the derivative of the following function from first principle. $\frac{1}{\text{x}^2}$
AnswerLet $\text{f}(\text{x})=\frac{1}{\text{x}^2}$. Accordingly,from the first principle, $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1}{(\text{x}+\text{h})^2}-\frac{1}{\text{x}^2}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{\text{x}^2-(\text{x}+\text{h})^2}{\text{x}^2(\text{x}+\text{h})^2}\Bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{\text{x}^2-\text{x}^2-\text{h}^2-2\text{hx}}{\text{x}^2(\text{x}+\text{h})^2}\Bigg]$$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{-\text{h}^2-2\text{hx}}{\text{x}^2(\text{x}+\text{h})^2}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{-\text{h}-2\text{x}}{\text{x}^2(\text{x}+\text{h})^2}\Bigg]$
$=\frac{0-2\text{x}}{\text{x}^2(\text{x}+0)^2}=\frac{-2}{\text{x}^3}$
View full question & answer→Question 104 Marks
Evaluate the following limits in Exercise:$\lim\limits_{\text{x}\rightarrow\pi}\frac{\sin(\pi-\text{x})}{\pi(\pi-\text{x})}$
Answer$\lim\limits_{\text{x}\rightarrow\pi}\frac{\sin(\pi-\text{x})}{\pi(\pi-\text{x})}$ It is seen that $\text{x} →\pi ⇒ (\pi –\text{x}) → 0$$\lim\limits_{\text{x}\rightarrow\pi}\frac{\sin(\pi-\text{x})}{\pi(\pi-\text{x})}=\frac{1}{\pi}\lim\limits_{(\text{x}-\pi)\rightarrow0}\frac{\sin(\pi-\text{x})}{(\pi-\text{x})}$
$=\frac{1}{\pi}\times1$ $\bigg[\lim\limits_{\text{y}\rightarrow0}\frac{\sin(\text{y})}{\text{y}}=1\bigg]$
$=\frac{1}{\pi}$
View full question & answer→Question 114 Marks
Find the derivative of the following function from first principle. $\frac{\text{x}+1}{\text{x}-1}$
AnswerLet $\text{f}(\text{x})=\frac{\text{x}+1}{\text{x}-1}$. Accordingly, from the first principle, $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\bigg(\frac{\text{x}+\text{h}+1}{\text{x}+\text{h}-1}-\frac{\text{x}+1}{\text{x}-1}\bigg)}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{(\text{x}-1)(\text{x}+\text{h}+1)-(\text{x}+1)(\text{x}+\text{h}-1)}{(\text{x}-1)(\text{x}+\text{h}-1)}\Bigg]$$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{(\text{x}^2+\text{hx}+\text{x}-\text{x}-\text{h}-1)(\text{x}^2+\text{hx}-\text{x}+\text{x}+\text{h}-1}{(\text{x}+1)(\text{x}+\text{h}-1)}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{-2\text{h}}{(\text{x}-1)(\text{x}+\text{h})-1}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{-2}{(\text{x}-1)(\text{x}+\text{h})-1}\Bigg]$ $=\frac{2}{(\text{x}-1)(\text{x}-1)}=\frac{-2}{(\text{x}-1)^2}$
View full question & answer→Question 124 Marks
Find the derivative of $x^2– 2$ at $x = 10$.
AnswerLet. $\text{f}(\text{x})=\text{x}^2-2$ Accordingly, $\text{f}'(10)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(10+\text{h})-\text{f}(10)}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\big[(10+\text{h})^2-2\big]-(10^2-2)}{\text{h}}$$=\lim\limits_{\text{h}\rightarrow0}\frac{10^2+2.10.\text{h}+\text{h}^2-2-10^2+2}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{20+\text{h}^2}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}(20+\text{h})=(20+0)=20$ Thus, the derivative of $x^2 – 2$ at $x = 10$ is $20$.
View full question & answer→Question 134 Marks
Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\frac{\text{a}}{\text{x}^4}-\frac{\text{b}}{\text{x}^2}+\cos\text{x}$
AnswerLet $\text{f(x)}=\frac{\text{a}}{\text{x}^4}-\frac{\text{b}}{\text{x}^2}+\cos\text{x}$ By quotient rule, $\text{f}'\text{(x)}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{a}}{\text{x}^4}\Big)-\frac{\text{d}}{\text{dx}}\Big(\frac{\text{b}}{\text{x}^2}\Big)+\frac{\text{d}}{\text{dx}}(\cos\text{x})$ $=\text{a}\frac{\text{d}}{\text{dx}}(\text{x}^{-4})-\text{b}\frac{\text{d}}{\text{dx}}(\text{x}^{-2})+\frac{\text{d}}{\text{dx}}(\cos\text{x})$ $=\text{a}(-4\text{x}^{-5})-\text{b}(-2\text{x}^{-3})+(-\sin\text{x})$ $\Big[\frac{\text{d}}{\text{dx}}(\text{x''})=\text{nx}^{\text{n}-1}\text{and}\frac{\text{d}}{\text{dx}}(\cos\text{x)}=-\sin\text{x}\Big]$ $=\frac{-4\text{a}}{\text{x}^5}+\frac{\text{2b}}{\text{x}^3}-\sin\text{x}$
View full question & answer→Question 144 Marks
Find the derivative of the following functions from first principle:$(-\text{x})^{-1}$
AnswerLet $\text{f(x)}=(-\text{x})^{-1}=\frac{1}{-\text{x}}=\frac{-1}{\text{x}}$. Accordingly, $\text{f(x+h)}=\frac{-1}{\text{(x+h)}}$ By first principle, $\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{-1}{\text{x+h}}-\Big(\frac{-1}{\text{x}}\Big)\Big]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{-1}{\text{x+h}}+\frac{1}{\text{x}}\Big]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{-\text{x}+(\text{x+h})}{\text{x(x+h)}}\Big]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{-\text{x+x+h}}{\text{x(x+h)}}\Big]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{\text{h}}{\text{x(x+h)}}\Big]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{x(x+h)}}$ $=\frac{1}{\text{x}.\text{x}}=\frac{1}{\text{x}^2}$
View full question & answer→Question 154 Marks
For some constants a and b, find the derivative of$(\text{a}\text{x}^2+\text{b})^2$
AnswerLet $\text{f}(\text{x})=(\text{a}\text{x}^2+\text{b})^2$$\Rightarrow\text{f}(\text{x})=\text{a}^2\text{x}^4+2\text{ab}\text{x}^2+\text{b}^2$
$\therefore\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}(\text{a}^2\text{x}^4+2\text{ab}\text{x}^2+\text{b}^2)=\text{a}^2\frac{\text{d}}{\text{dx}}(\text{x}^4)+2\text{ab}\frac{\text{d}}{\text{dx}}(\text{x}^2)+\frac{\text{d}}{\text{dx}}(\text{b}^2)$ On using theorem $\frac{\text{d}}{\text{dx}}(\text{x})^\text{n}=\text{nx}^{\text{n}-1}$, we obtain $\text{f}'(\text{x})=\text{a}^2(4\text{x}^3)+2\text{ab}(2\text{x})+\text{b}^2(0)$ $=4\text{a}^2\text{x}^3+4\text{ab}\text{x}$ $=4\text{a}\text{x}(\text{a}\text{x}^2+\text{b})$
View full question & answer→Question 164 Marks
Evaluate the following limits in Exercise:$\lim\limits_{\text{x}\rightarrow0}(\text{cosec x}-\cot\text{x})$
AnswerAt x = 0, the value of the given function takes the form$\infty-\infty$. Now, $\lim\limits_{\text{x}\rightarrow0}(\text{cosec x}-\cot\text{x})$ $=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}\Big)$ $=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{1-\cos\text{x}}{\sin\text{x}}\Big)$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\Big(\frac{1-\cos\text{x}}{\text{x}}\Big)}{\Big(\frac{\sin\text{x}}{\text{x}}\Big)}$ $=\frac{\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos\text{x}}{\text{x}}}{\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}}$ $=\frac{0}{1}$ $\bigg[{\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos\text{x}}{\text{x}}=0}\text{and}{\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}}=1\bigg]$ $=0$
View full question & answer→Question 174 Marks
If the function f(x) satisfies $\lim\limits_{\text{x}\rightarrow1}\frac{\text{f}(\text{x})-2}{\text{x}^2-1}=\pi$ , evaluate $\lim\limits_{\text{x}\rightarrow1}{\text{f}(\text{x})}$.
Answer$\lim\limits_{\text{x}\rightarrow1}\frac{\text{f}(\text{x})-2}{\text{x}^2-1}=\pi$$\Rightarrow\frac{\lim\limits_{\text{x}\rightarrow1}{(\text{f}(\text{x})-2)}}{\lim\limits_{\text{x}\rightarrow1}(\text{x}^2-1)}=\pi$
$\Rightarrow{\lim\limits_{\text{x}\rightarrow1}{(\text{f}(\text{x})-2)}}=\pi{\lim\limits_{\text{x}\rightarrow1}(\text{x}^2-1)}$
$\Rightarrow{\lim\limits_{\text{x}\rightarrow1}{(\text{f}(\text{x})-2)}}=0$
$\Rightarrow{\lim\limits_{\text{x}\rightarrow1}{\text{f}(\text{x})}}-{\lim\limits_{\text{x}\rightarrow1}2}=0$
$\Rightarrow{\lim\limits_{\text{x}\rightarrow1}{\text{f}(\text{x})-2}}=0$
$\therefore{\lim\limits_{\text{x}\rightarrow1}{\text{f}(\text{x})}}=2$
View full question & answer→Question 184 Marks
Evaluate the following limits in Exercise:$\lim\limits_{\text{z}\rightarrow1}\frac{\text{z}^{\frac{1}{3}}-1}{\text{z}^{\frac{1}{6}-1}}$
Answer$\lim\limits_{\text{z}\rightarrow1}\frac{\text{z}^{\frac{1}{3}}-1}{\text{z}^{\frac{1}{6}-1}}$At z = 1, the value of the given function takes the form $\frac{0}{0}$.
Put $\text{z}^\frac{1}{6}=\text{x}$ so that z →1 as x → 1.
$\lim\limits_{\text{z}\rightarrow1}\frac{\text{z}^{\frac{1}{3}}-1}{\text{z}^{\frac{1}{6}-1}}$ $=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^2-1^2}{\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^2-1^2}{\text{x}-1} $
$=2.1^{2.1} $ $\Big[\lim\limits_{\text{x}\rightarrow\text{a}}\frac{\text{x}^\text{n}-\text{a}^\text{n}}{\text{x}-\text{a}}=\text{na} ^{\text{n}-1}\Big]$
$=2$
$\therefore\lim\limits_{\text{z}\rightarrow1}\frac{\text{z}^{\frac{1}{3}}-1}{\text{z}^{\frac{1}{6}-1}}=2$
View full question & answer→Question 194 Marks
Find $\lim\limits_{\text{x}\rightarrow1}\text{f}(\text{x})$, where $\text{f}(\text{x})=\begin{cases}\text{x}^2-1,& x \leq 1\\-\text{x}^2-1, & x > 1\end{cases}$
AnswerThe given function is$\text{f}(\text{x})=\begin{cases}\text{x}^2-1,& x \leq 1\\-\text{x}^2-1, & x > 1\end{cases}$
$\lim\limits_{\text{x}\rightarrow1^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow1}[\text{x}^2-1]=1^2-1=1-1=0$ $\lim\limits_{\text{x}\rightarrow1^+}\text{f}(-\text{x})=\lim\limits_{\text{x}\rightarrow1}[\text{x}^2-1]=-1^2-1=-1-1=-2$ It is observed that $\lim\limits_{\text{x}\rightarrow1^-}\text{f}(\text{x})\neq\lim\limits_{\text{x}\rightarrow1^+}\text{f}(\text{x})$. Hence, $\lim\limits_{\text{x}\rightarrow1}\text{f}(\text{x})$ does not exist.
View full question & answer→Question 204 Marks
Evaluate the following limits in Exercise:$\lim\limits_{\text{x}\rightarrow0}\frac{\sin{\text{ax}}}{\text{bx}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\sin{\text{ax}}}{\text{bx}}$At x = 0, the value of the given function takes the form $\frac{0}{0}$.
$\text{Now},\lim\limits_{\text{x}\rightarrow0}\frac{\sin{\text{ax}}}{\text{bx}}=\lim\limits_{\text{x}\rightarrow0}\frac{\sin{\text{ax}}}{\text{ax}}\times\frac{\text{ax}}{\text{bx}}$
$=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\sin{\text{ax}}}{\text{ax}}\Big)\times\Big(\frac{\text{a}}{\text{b}}\Big)$
$=\frac{\text{a}}{\text{b}}\lim\limits_{\text{ax}\rightarrow0}\Big(\frac{\sin{\text{ax}}}{\text{ax}}\Big)$ $[\text{x}\rightarrow0\Rightarrow\text{ax}\rightarrow0]$
$=\frac{\text{a}}{\text{b}}\times1$ $\Big[\lim\limits_{\text{y}\rightarrow0}\frac{\sin{\text{y}}}{\text{y}}=1\Big]$
$=\frac{\text{a}}{\text{b}}$
View full question & answer→Question 214 Marks
Find the derivative of the following function from first principle. $\text{x}^3-27$
AnswerLet $\text{f}(\text{x})=\text{x}^3-27$. Accordingly,from the first principle, $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\big[(\text{x}+\text{h})^3-27\big]-(\text{x}^3-27)}{\text{h}}$ $\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}^3+3\text{x}^2\text{h}+3\text{x}\text{h}^2}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}{(\text{h}^2+3\text{x}^2+3\text{x}\text{h})}$$=0+3\text{x}^2+0=3\text{x}^2$
View full question & answer→Question 224 Marks
Evaluate the following limits in Exercise:$\lim\limits_{\text{x}\rightarrow0}\frac{\text{ax}+\text{x}\cos\text{x}}{\text{b}\sin\text{x}}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\text{ax}+\text{x}\cos\text{x}}{\text{b}\sin\text{x}}$At x = 0, the value of the given function takes the form$\frac{0}{0}$.
$\text{Now},$
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{ax}+\text{x}\cos\text{x}}{\text{b}\sin\text{x}}=\frac{1}{\text{b}}\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}(\text{a}+\cos\text{x})}{\sin\text{x}}$
$=\frac{1}{\text{b}}\lim\limits_{\text{x}\rightarrow0}\bigg(\frac{\text{x}}{\sin\text{x}}\bigg)\times\lim\limits_{\text{x}\rightarrow0}(\text{a}+\cos\text{x})$
$=\frac{1}{\text{b}}\times\bigg(\frac{1}{\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}}\bigg) \times\lim\limits_{\text{x}\rightarrow0}{(\text{a}+\cos\text{x})}$
$=\frac{1}{\text{b}}\times(\text{a}+\cos0) \bigg[\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\bigg]$
$=\frac{\text{a}+1}{\text{b}}$
View full question & answer→Question 234 Marks
If $\text{f}(\text{x})=\begin{cases}a & x = 0\\b & x > 0\end{cases}\begin{cases}\text{mx}^2+\text{n,} & x < 0\\\text{nx}+m, & 0\leq\text{x}\leq 1\\\text{nx}^3+\text{x}&\text{x}>1\end{cases}$ . For what integers m and n does both $\lim\limits_{\text{x}\rightarrow0}\text{f}(\text{x})$ and $\lim\limits_{\text{x}\rightarrow1}\text{f}(\text{x})$exist?
AnswerThe given function is$\text{f}(\text{x})=\begin{cases}a & x = 0\\b & x > 0\end{cases}\begin{cases}\text{mx}^2+\text{n,} & x < 0\\\text{nx}+m, & 0\leq\text{x}\leq 1\\\text{nx}^3+\text{x}&\text{x}>1\end{cases}$
$\lim\limits_{\text{x}\rightarrow0^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow0}(\text{mx}^2+\text{n})$
$=\text{m}(0)^2+\text{n}$
$=\text{n}$ $\lim\limits_{\text{x}\rightarrow0^+}=\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow0}(\text{nx}+\text{m})$ $=\text{n}(0)+\text{m}$ $=\text{m}$. Thus, $\lim\limits_{\text{x}\rightarrow0}\text{f}(\text{x})$exists if m = n . $\lim\limits_{\text{x}\rightarrow1^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow1}(\text{nx}+\text{m})$ $=\text{n}(1)+\text{m}$$=\text{m+n}$
$\lim\limits_{\text{x}\rightarrow1^+}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow1}(\text{nx}^3+\text{m})$ $=\text{n}(1)^3+\text{m}$ $=\text{m+n}$ $\therefore\lim\limits_{\text{x}\rightarrow1^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow1^+}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow1}\text{f}(\text{x})$. Thus, $\lim\limits_{\text{x}\rightarrow1}\text{f}(\text{x})$exists for any integral value of m and n.
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Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $(\text{ax+b})(\text{cx+d})^2$
AnswerLet $\text{f(x)}=\text{(ax+b)(cx+d)}^2$ By Leibnitz product rule, $\text{f}'\text{(x)}=\text{(ax+b)}\frac{\text{d}}{\text{dx}}\text{(cx+d)}^2+\text{(cx+d)}^2\frac{\text{d}}{\text{dx}}\text{(ax+b)}$ $=\text{(ax+b)}\frac{\text{d}}{\text{dx}}(\text{c}^2\text{x}^2+\text{2cdx+d}^2)+\text{cx+d}^2\frac{\text{d}}{\text{dx}}\text{(ax+b)}$ $=\text{(ax+b)}\Big[\frac{\text{d}}{\text{dx}}(\text{c}^2\text{x}^2)\frac{\text{d}}{\text{dx}}+(2\text{cdx})+\frac{\text{d}}{\text{dx}}\text{d}^2\Big]+(\text{cx+d})^2\Big[\frac{\text{d}}{\text{dx}}\text{ax+}\frac{\text{d}}{\text{dx}}\text{b}\Big]$ $=\text{(ax+b)}(\text{2c}^2\text{x}+2\text{cd})+(\text{cx+d}^2)\text{a}$ $=2\text{c}(\text{ax+b)(cx+d})+\text{a(cx+d)}^2$
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Find the derivative of 99x at x = l00 .
AnswerLet $\text{f}(\text{x})=\text{x}$. Accordingly, $\text{f}'(1)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-\text{f}(1)}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{(1+\text{h})-1}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}(1)$$=1$
Thus, the derivative of at x at x = 1 is 1.
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Find the derivative of the following functions from first principle:$\sin(\text{x+1})$
AnswerLet $\text{f(x)}=\sin(\text{x+1})$. Accordingly, $\text{f}\text{(x+1)}=\sin(\text{x+h+1})$ By first principle, $\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}[\sin(\text{x+h+1})-\sin(\text{x+1})]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[2\cos\Big (\frac{\text{x+h+1+x+1}}{2}\Big)\sin\Big(\frac{\text{x+h+1}-\text{x}-1}{2}\Big)\Big]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[2\cos\Big(\frac{2\text{x}+\text{h}+2}{2}\Big)\sin\Big(\frac{\text{h}}{2}\Big)\Big]$ $=\lim\limits_{\text{h}\rightarrow0}\Bigg[\cos\Big(\frac{2\text{x+h+2}}{2}\Big).\frac{\sin\Big(\frac{\text{h}}{2}\Big)}{\Big(\frac{\text{h}}{2}\Big)}\Bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\cos\Big(\frac{\text{2x+h+2}}{2}\Big).\lim\limits_{\frac{\text{h}}{2}\rightarrow0}\frac{\sin\Big(\frac{\text{h}}{2}\Big)}{\Big(\frac{\text{h}}{2}\Big)}\ $$\Big[\text{As h}\rightarrow0\Rightarrow\frac{\text{h}}{2}\rightarrow0\Big]$ $=\cos\Big(\frac{\text{2x+0+2}}{2}\Big).1$ $\Big[\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$ $=\cos(\text{x}+1)$
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Evaluate the following limits in Exercise:$\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^4-81}{2\text{x}^2-5\text{x-3}}$
AnswerAt x = 2, the value of the given rational function takes the form $\frac{0}{0}$.$\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}^4-81}{2\text{x}^2-5\text{x-3}}$$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}-3)(\text{x}+3)(\text{x}^2+9)}{(\text{x}-3)(2\text{x}+1)}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}+3)(\text{x}^2+9)}{2\text{x}+1}$
$=\frac{(3+3)(3^2+9)}{2(3)+1}$
$=\frac{6\times18}{7}$
$=\frac{108}{7}$
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Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\frac{\text{px}^2\text{+qx+r}}{\text{ax+b}}$
AnswerLet $\text{f(x)}=\frac{\text{px}^2\text{+qx+r}}{\text{ax+b}}$ By quotient rule, $\text{f}'\text{(x)}=\frac{(\text{ax+b)}\frac{\text{d}}{\text{dx}}(\text{px}^2\text{+qx+r})-(\text{px}^2\text{+qx+r})\frac{\text{d}}{\text{dx}}(\text{ax+b})}{(\text{ax+b})^2}$ $=\frac{(\text{ax+b})(\text{2px+q})-(\text{px}^2\text{qx+r)(a)}}{(\text{ax+b})^2}$ $=\frac{\text{2apx}^2+\text{aqx}+2\text{bpx}+\text{bq}+\text{apx}^2-\text{aqx}-\text{ar}}{(\text{ax+b})^2}$ $=\frac{\text{apx}^2-2\text{bpx+bq}-\text{ar}}{(\text{ax+b})^2}$
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Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\frac{\text{x}^2\cos\Big(\frac{\pi}{4}\Big)}{\sin\text{x}}$
AnswerLet $\text{f(x)}=\frac{\text{x}^2\cos\Big(\frac{\pi}{4}\Big)}{\sin\text{x}}$ By quotient rule, $\text{f}'\text{(x)}=\cos\frac{\pi}{4}.\Bigg[\frac{\sin\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^2)-\text{x}^2\frac{\text{d}}{\text{dx}}(\sin\text{x})}{\sin^2\text{x}}\Bigg]$ $=\cos\frac{\pi}{4}.\Big[\frac{\sin\text{x}.2\text{x}-\text{x}^2\cos\text{x}}{\sin^2\text{x}}\Big]$ $=\frac{\text{x}\cos\frac{\pi}{4}[2\sin\text{x}-\text{x}\cos\text{x}]}{\sin^2\text{x}}$
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Suppose $\text{f}(\text{x})=\begin{cases}\text{a+bx},& \text{x}< 0\\4 & \text{x} = 1\\\text{b-ax},&\text{x}>1\end{cases}$and if $\lim\limits{\text{x}\rightarrow1}=\text{f}(1)$ what are possible values of a and b?
AnswerThe given function is $\text{f}(\text{x})=\begin{cases}\text{a+bx},& \text{x}< 0\\4 & \text{x} = 1\\\text{b-ax},&\text{x}>1\end{cases}$ $=\lim\limits_{\text{x}\rightarrow1^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow1}(\text{a+bx})=\text{a+b}$ $=\lim\limits_{\text{x}\rightarrow1^+}\text{f}(\text{x})==\lim\limits_{\text{x}\rightarrow1}(\text{b-ax})=\text{b-a}$ $=\text{f}(1)=4$It is given that $\lim\limits_{\text{1}\rightarrow1}\text{f}(\text{x})=\text{f}(1)$
$\therefore=\lim\limits_{\text{x}\rightarrow1^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow1^+}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow1}\text{f}(\text{x})=\text{f}(1)$
$\Rightarrow\text{a+b}=4$and$\text{b-a}=4$ On solving these two equations, we obtain a=0 nd b=4. Thus, the respective possible values of a and b are 0 and 4.
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Find the derivative of $\frac{\text{x}^\text{n}-\text{a}^\text{n}}{\text{x}-\text{a}}$ for some constant a.
AnswerLet $\text{f}(\text{x})=\frac{\text{x}^\text{n}-\text{a}^\text{n}}{\text{x}-\text{a}}$ $\Rightarrow\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}\bigg(\frac{\text{x}^\text{n}-\text{a}^\text{n}}{\text{x}-\text{a}}\bigg)$ By quotient rule, $ \text{f}'(\text{x})=\frac{({\text{x}-\text{a})\frac{\text{d}}{\text{dx}}(\text{x}^\text{n}-\text{a}^\text{n})-(\text{x}^\text{n}-\text{a}^\text{n})}\frac{\text{d}}{\text{dx}}(\text{x}-\text{a})}{(\text{x}-\text{a})^2}$ $ =\frac{({\text{x}-\text{a})(\text{nx}^\text{n-1}-0)-(\text{x}^\text{n}-\text{a}^\text{n})}}{(\text{x}-\text{a})^2}$ $ =\frac{{\text{nx}^\text{n}-\text{anx}^\text{n-1}-\text{x}^\text{n}+\text{a}^\text{n}}}{(\text{x}-\text{a})^2}$
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For some constants a and b, find the derivative of$(\text{x}-\text{a})(\text{x}-\text{b})$
AnswerLet $\text{f}(\text{x})=(\text{x}-\text{a})(\text{x}-\text{b})$$\Rightarrow\text{f}(\text{x})=\text{x}^2-(\text{a}+\text{b})\text{x}+\text{ab}$
$\therefore\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}(\text{x}^2-(\text{a}+\text{b})\text{x}+\text{ab})$$=\frac{\text{d}}{\text{dx}}(\text{x}^2)-(\text{a}+\text{b})\frac{\text{d}}{\text{dx}}(\text{x})+\frac{\text{d}}{\text{dx}}(\text{ab})$
On using theorem $\frac{\text{d}}{\text{dx}}(\text{x})^\text{n}=\text{nx}^{\text{n}-1}$, we obtain $\text{f}'(\text{x})=2\text{x}-(\text{a}+\text{b})+0=2\text{x}-\text{a}-\text{b}$
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Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\frac{\text{ax+b}}{\text{cx+d}}$
AnswerLet $\text{f(x)}=\frac{\text{ax+b}}{\text{cx+d}}$ By quotient rule, $\text{f}'\text{(x)}=\frac{\text{(cx+d)}\frac{\text{d}}{\text{dx}}\text{(ax+b)}-\text{(ax+b)}\frac{\text{d}}{\text{dx}}\text{(cx+d)}}{(\text{cx+d})^2}$ $=\frac{\text{(cx+d)(a)}-\text{(ax+b)}\text{(c)}}{\text{(cx+d)}^2}$ $=\frac{\text{acx+ad}-\text{acx+bc}}{\text{(cx+d})^2}$ $=\frac{\text{ad}-\text{bc}}{\text{(cx+d)}^2}$
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Find the derivative of x at x = 1.
AnswerLet $\text{f}(\text{x})=99\text{x}$. Accordingly, $\text{f}'(100)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(100+\text{h})-\text{f}(100)}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{99(100+\text{h})-99(100)}{\text{h}}$$=\lim\limits_{\text{h}\rightarrow0}\frac{99\times100+99\text{h}-99\times100}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{99\text{h}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}(99)=99$ Thus, the derivative of at x = 100 is 99.
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Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}}$
AnswerLet $\text{f(x)}=\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}}$ By quotient rule, $\text{f}'\text{(x)}=\frac{(\sin\text{x}-\cos\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x}+\cos\text{x})-(\sin\text{x}+\cos\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x}-\cos\text{x})}{(\sin\text{x}-\cos\text{x})^2}$ $=\frac{(\sin\text{x}-\cos\text{x})(\cos\text{x}-\sin\text{x})-(\sin\text{x}+\cos\text{x})(\cos\text{x}+\sin\text{x})}{(\sin\text{x}-\cos\text{x})^2}$ $=\frac{-(\sin\text{x}-\cos\text{x})^2-(\sin\text{x}+\cos\text{x})^2}{(\sin\text{x}-\cos\text{x})^2}$ $=\frac{\big[\sin^2\text{x}+\cos^2\text{x}-2\sin\text{x}\cos\text{x}+\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x}\big]}{(\sin\text{x}-\cos\text{x})^2}$ $=\frac{[1+1]}{(\sin\text{x}-\cos\text{x})^2}$ $=\frac{-2}{(\sin\text{x}-\cos\text{x})^2}$
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Find $\lim\limits_{\text{x}\rightarrow0}\text{f}(\text{x})$, where $\text{f}(\text{x})\begin{cases}\frac{|\text{x}|}{\text{x}}, &\text{x}\neq 0\\0, & x > 0\end{cases}$
AnswerThe given function is $\text{f}(\text{x})\begin{cases}\frac{|\text{x}|}{\text{x}}, &\text{x}\neq 0\\0, & x > 0\end{cases}$ $\lim\limits_{\text{x}\rightarrow0^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow0^-}\big[\frac{|\text{x}|}{\text{x}}\big]$ $=\lim\limits_{\text{x}\rightarrow0}\big(\frac{-\text{x}}{\text{x}}\big)$ [When x is negative,$|\text{x}|=-\text{x}$] $=\lim\limits_{\text{x}\rightarrow0}(-1)$ $=-1$$\lim\limits_{\text{x}\rightarrow0^+}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow0^+}\big[\frac{|\text{x}|}{\text{x}}\big]$
$=\lim\limits_{\text{x}\rightarrow0}\big[\frac{\text{x}}{\text{x}}\big]$ [When x is positive,$|\text{x}|=\text{x}$] $=\lim\limits_{\text{x}\rightarrow0}(1)$ $=1$ It is observed that $\lim\limits_{\text{x}\rightarrow0^-}\text{f}(\text{x})\neq\lim\limits_{\text{x}\rightarrow0^+}\text{f}(\text{x})$ . Hence, $\lim\limits_{\text{x}\rightarrow0}\text{f}(\text{x})$ does not exist.
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Find the derivative of. $\frac{2}{\text{x}-1}-\frac{\text{x}^2}{3\text{x}-1}$
AnswerLet $\text{f}(\text{x})=\frac{2}{\text{x}-1}-\frac{\text{x}^2}{3\text{x}-1}$ $\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}\bigg(\frac{2}{\text{x}+1}\bigg)-\frac{\text{d}}{\text{dx}}\bigg(\frac{\text{x}^2}{3\text{x}-1}\bigg)$ By quotient rule, $\text{f}'\big(\text{x})=\Bigg[\frac{(\text{x}+1)\frac{\text{d}}{\text{dx}}(2)-2\frac{\text{d}}{\text{dx}}(\text{x}+1)}{(\text{x}+1)^2}\Bigg]-\Bigg[\frac{(3\text{x}-1)\frac{\text{d}}{\text{dx}}(\text{x}^2)-\text{x}^2\frac{\text{d}}{\text{dx}}(3\text{x}-1)}{(3\text{x}+1)^2}\Bigg]$$=\bigg[\frac{(\text{x}+1)(0)-2(1)}{(\text{x}+1)^2}\bigg]-\bigg[\frac{(3\text{x}-1)(2\text{x})-(\text{x}^1)(3)}{(3\text{x}-1)^2}\bigg]$
$=\frac{-2}{(\text{x}+1)^2}-\bigg[\frac{6\text{x}^2-2\text{x}-3\text{x}^2}{(3\text{x}-1)^2}\bigg]$
$=\frac{-2}{(\text{x}+1)^2}-\bigg[\frac{3\text{x}^2-2\text{x}^2}{(3\text{x}-1)^2}\bigg]$
$=\frac{2}{(\text{x}+1)^2}-\frac{\text{x}(3\text{x-2})}{(3\text{x}-1)^2}$
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Evaluate the following limits in Exercise:$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{\tan2\text{x}}{\text{x}-\frac{\pi}{2}}$
Answer$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{\tan2\text{x}}{\text{x}-\frac{\pi}{2}}$ At $\text{x}=\frac{\pi}{2}$, the value of the given function takes the form $\frac{0}{0}$. Now, put $\text{x}-\frac{\pi}{2}=\text{y}$ so that $\text{x}\rightarrow\frac{\pi}{2},\text{y}\rightarrow0$. $\therefore\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{\tan2\text{x}}{\text{x}-\frac{\pi}{2}}=\lim\limits_{\text{y}\rightarrow0}\frac{\tan2\big(\text{y}+\frac{\pi}{2}\big)}{\text{y}}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{\tan(\pi+2\text{y})}{\text{y}}$ $=\lim\limits_{\text{y}\rightarrow0}\frac{\tan2\text{y}}{\text{y}}$ $[\tan(\pi+2\text{y})=\tan2\text{y}]$ $=\lim\limits_{\text{y}\rightarrow0}\frac{\sin2\text{y}}{\text{y}\cos2\text{y}}$ $=\lim\limits_{\text{y}\rightarrow0}\bigg(\frac{\sin2\text{y}}{2\text{y}}\times\frac{2}{\cos2\text{y}}\bigg)$ $=\bigg(\lim\limits_{2\text{y}\rightarrow0}\frac{\sin2\text{y}}{2\text{y}}\bigg)\times\lim\limits_{\text{y}\rightarrow0}\bigg(\frac{2}{\cos2\text{y}}\bigg)$ $[\text{y}\rightarrow0\Rightarrow2\text{y}\rightarrow0]$ $=1\times\frac{2}{\cos0}$ $\big[=\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\big]$$=1\times\frac{2}{1}$
$=2$
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For some constants a and b, find the derivative of$\frac{\text{x}-\text{a}}{\text{x}-\text{b}}$
Answer$\Rightarrow\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}\bigg(\frac{\text{x}-\text{a}}{\text{x}-\text{b}}\bigg)$ By quotient rule, $\text{f}'(\text{x})=\frac{(\text{x}-\text{b})\frac{\text{d}}{\text{dx}}(\text{x}-\text{a})-(\text{x}-\text{a})\frac{\text{d}}{\text{dx}}(\text{x}-\text{b})}{(\text{x}-\text{b})^2}$ $=\frac{\text{x}-\text{b}-\text{x}+\text{a}}{(\text{x}-\text{b})^2}$$=\frac{\text{a}-\text{b}}{(\text{x}-\text{b})^2}$
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Find $\lim\limits_{\text{x}\rightarrow0}\text{f}(\text{x})$, where $\text{f}(\text{x})\begin{cases}\frac{\text{x}}{|\text{x}|}, &\text{x}\neq 0\\0, & \text{x} = 0\end{cases}$
AnswerThe given function is $\text{f}(\text{x})\begin{cases}\frac{\text{x}}{|\text{x}|}, &\text{x}\neq 0\\0, & \text{x} = 0\end{cases}$ $\lim\limits_{\text{x}\rightarrow0^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow0^-}\big[\frac{\text{x}}{|\text{x}|}\big]$ $=\lim\limits_{\text{x}\rightarrow0}\big(\frac{\text{x}}{\text{-x}}\big)$ [When x<0,$|\text{x}|=-\text{x}$] $=\lim\limits_{\text{x}\rightarrow0}(-1)$ $=-1$$\lim\limits_{\text{x}\rightarrow0^+}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow0^+}\big[\frac{\text{x}}{|\text{x}|}\big]$
$=\lim\limits_{\text{x}\rightarrow0}\big[\frac{\text{x}}{\text{x}}\big]$ [When x>0,$|\text{x}|=\text{x}$] $=\lim\limits_{\text{x}\rightarrow0}(1)$ $=1$ It is observed that$\lim\limits_{\text{x}\rightarrow0^-}\text{f}(\text{x})\neq\lim\limits_{\text{x}\rightarrow0^+}\text{f}(\text{x})$ . Hence, $\lim\limits_{\text{x}\rightarrow0}\text{f}(\text{x})$ does not exist.
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Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\frac{\text{ax+b}}{\text{px}^2+\text{qx+r}}$
AnswerLet $\text{f(x)}=\frac{\text{ax+b}}{\text{px}^2+\text{qx+r}}$ By quotient rule, $\text{f}'\text{(x)}=\frac{(\text{px}^2\text{+qx+r})\frac{\text{d}}{\text{dx}}(\text{ax+b)}-(\text{ax+b})\frac{\text{d}}{\text{dx}}(\text{px}^2\text{+qx+r})}{(\text{px}^2\text{+qx+r})^2}$ $=\frac{(\text{px}^2\text{+qx+r})(\text{a})-(\text{ax+b})(2\text{px+q)}}{(\text{px}^2\text{+qx+r})^2}$ $=\frac{\text{apx}^2+\text{aqx}+\text{ar}-2\text{apx}^2-\text{aqx}-2\text{bpx}-\text{bq}}{(\text{px}^2\text{+qx+r})^2}$ $=\frac{-\text{apx}^2-2\text{bpx+ar}-\text{bq}}{(\text{px}^2\text{+qx+r})^2}$
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Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\frac{\sec\text{x}-1}{\sec\text{x+1}}$
AnswerLet $\text{f(x)}=\frac{\sec\text{x}-1}{\sec\text{x+1}}$ $\text{f(x)}=\frac{\frac{1}{\cos\text{x}}-1}{\frac{1}{\cos\text{x}}+1}=\frac{1-\cos\text{x}}{1+\cos\text{x}}$ By quotient rule, $\text{f}'\text{(x)}=\frac{(1+\cos\text{x})\frac{\text{d}}{\text{dx}}(1-\cos\text{x})-(1-\cos\text{x})\frac{\text{d}}{\text{dx}}(1+\cos\text{x})}{(1+\cos\text{x})^2}$ $=\frac{(1+\cos\text{x})(\sin\text{x})-(1-\cos\text{x})(-\sin\text{x})}{(1+\cos\text{x})^2}$ $=\frac{\sin\text{x}+\cos\text{x}\sin\text{x}+\sin\text{x}-\sin\text{x}\cos\text{x}}{(1+\cos\text{x})^2}$ $=\frac{2\sin\text{x}}{(1+\cos\text{x})^2}$ $=\frac{2\sin\text{x}}{\Big(1+\frac{1}{\sec\text{x}}\Big)^2}$ $=\frac{2\sin\text{x}}{\frac{(\sec\text{x}+1)^2}{\sec^2\text{x}}}$$=\frac{2\sin\text{x}\sec^2\text{x}}{(\sec\text{x+1})^2}$
$=\frac{\frac{2\sin\text{x}}{\cos\text{x}}\sec\text{x}}{(\sec\text{x+1})^2}$ $=\frac{2\sec\text{x}\tan\text{x}}{(\sec\text{x+1})^2}$
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Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $(\text{x}+\sec\text{x})(\text{x}-\tan\text{x})$
AnswerLet $\text{f(x)}=(\text{x}+\sec\text{x})(\text{x}-\tan\text{x})$By product rule,
$\text{f}'\text{(x)}=(\text{x}+\sec\text{x})\frac{\text{d}}{\text{dx}}(\text{x}-\tan\text{x})(\text{x}-\tan\text{x})\frac{\text{d}}{\text{dx}}(\text{x}+\sec\text{x})$
$={(\text{x}+\sec\text{x})\Big[\frac{\text{d}}{\text{dx}}(\text{x})}-\frac{\text{d}}{\text{dx}}\tan\text{x}\Big]+(\text{x}-\tan\text{x})\Big[\frac{\text{d}}{\text{dx}}(\text{x})+\frac{\text{d}}{\text{dx}}\sec\text{x}\Big]$
$={(\text{x}+\sec\text{x})\Big[1-\frac{\text{d}}{\text{dx}}}\tan\text{x}\Big]+(\text{x}-\tan\text{x})\Big[1+\frac{\text{d}}{\text{dx}}\sec\text{x}\Big]$
$\text{Let f}_1\text{(x)}=\tan\text{x},$$\text{ f}_2\text{(x)}=\sec\text{x}$
$\text{Accordingly, f}_1(\text{x+h})=\tan(\text{x+h})\text{ and }$$\text{f}_2(\text{x+h})=\sec(\text{x+h})$
$\text{f}'_1\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\Big[\frac{\text{f}_1\text{(x+h)}-\text{f}_1\text{(x)}}{\text{h}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\tan(\text{x+h)}-\tan\text{x}}{\text{h}}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\Big[\frac{\tan(\text{x+h)}-\tan\text{x}}{\text{h}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{\sin(\text{x+h)}}{\cos(\text{x+h)}}-\frac{\sin\text{x}}{\cos\text{x}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{\sin(\text{x+h}-\text{x})}{\cos(\text{x+h)}\cos\text{x}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{\sin\text{h}}{\cos(\text{x+h})\cos\text{x}}\Big]$
$=\Big(\lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}\Big).\Big(\lim\limits_{\text{h}\rightarrow0}\frac{1}{\cos(\text{x+h})\cos\text{x}}\Big)$
$=1\times\frac{1}{\cos^2\text{x}}=\sec^2\text{x}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\tan\text{x}=\sec^2\text{x}\ ...(\text{ii})$
$\text{f}'_2\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\Big[\frac{\text{f}_2\text{(x+h)}-\text{f}_2\text{(x})}{\text{h}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\sec(\text{x+h)}-\sec\text{x}}{\text{h}}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{1}{\cos(\text{x+h)}}-\frac{1}{\cos\text{x}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{\cos\text{x}-\cos\text{(x+h)}}{\cos(\text{x+h)}\cos\text{x}}\Big]$
$=\frac{1}{\cos\text{x}}.\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{-2\sin\big(\frac{\text{x+x+h}}{2}\big).\sin\big(\frac{\text{x}-\text{x}-\text{h}}{2}\big)}{\cos(\text{x+h})}\Bigg]$
$=\frac{1}{\cos\text{x}}.\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{-2\sin\big(\frac{\text{2x+h}}{2}\big).\sin\big(\frac{-\text{h}}{2}\big)}{\cos(\text{x+h})}\Bigg]$
$=\frac{1}{\cos\text{x}}.\lim\limits_{\text{h}\rightarrow0}\Bigg[\frac{\sin\big(\frac{\text{2x+h}}{2}\big)\Big\{\frac{\sin\big(\frac{\text{h}}{2}\big)}{\frac{\text{h}}{2}}\Big\}}{\cos(\text{x+h})}\Bigg]$
$=\sec\text{x.}\frac{\Big\{\lim\limits_{\text{h}\rightarrow0}\sin\Big(\frac{2\text{x+h}}{2}\Big)\Big\}\Bigg\{\lim\limits_{\frac{\text{h}}{2}\rightarrow0}\frac{\sin\big(\frac{\text{h}}{2}\big)}{\frac{\text{h}}2}\Bigg\}}{\lim\limits_{\text{h}\rightarrow0}\cos(\text{x+h})}$
$=\sec\text{x.}\frac{\sin\text{x}.1}{\cos\text{x}}$
$\Rightarrow\frac{\text{d}}{\text{dx}}\sec\text{x}=\sec\text{x}\tan\text{x}\ ...(\text{iii})$
From (i), (ii), and (iii), we obtain
$\text{f}'(\text{x})=(\text{x}+\sec\text{x})(1-\sec^2\text{x})+(\text{x}-\tan\text{x})(1+\sec\text{x}\tan\text{x})$
View full question & answer→Question 444 Marks
If $\text{f}(\text{x})=\begin{cases}|\text{x}|+1,& \text{x}< 0\\0 & \text{x} = 0\\|\text{x}|-1,&\text{x}>1\end{cases}$. For what value (s) of a does $\lim\limits_{\text{x}\rightarrow\text{a}}\text{f}(\text{x})$ exists?
AnswerThe given function is $\text{f}(\text{x})=\begin{cases}|\text{x}|+1,& \text{x}< 0\\0 & \text{x} = 0\\|\text{x}|-1,&\text{x}>1\end{cases}$ When a=0, $\lim\limits_{\text{x}\rightarrow0^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow0^-}(|\text{x}|+1)$ $=\lim\limits_{\text{x}\rightarrow0}(-\text{x}+1)$ [If x<0,$|\text{x}|=-\text{x}$]$=-0+1$
$=1$ $\lim\limits_{\text{x}\rightarrow0^+}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow0^+}(|\text{x}|-1)$$=\lim\limits_{\text{x}\rightarrow0}(-\text{x}+1)$ [If x>0,$|\text{x}|=\text{x}$]
$=0-1$
$=-1$Here,it is observed that$\lim\limits_{\text{x}\rightarrow0^-}\text{f}(\text{x})\neq\lim\limits_{\text{x}\rightarrow0^+}\text{f}(\text{x})$.
$\therefore\lim\limits_{\text{x}\rightarrow0}\text{f}(\text{x})$does not exist.
When a < 0,$\lim\limits_{\text{x}\rightarrow\text{a}^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow\text{a}^-}(|\text{x}|+1)$
$=\lim\limits_{\text{x}\rightarrow\text{a}}(-\text{x}+1)$ [$\text{x}<\text{a}<0\Rightarrow|\text{x}|=-\text{x}$]
$=-\text{a}+1$
$\therefore\lim\limits_{\text{x}\rightarrow\text{a}^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow\text{a}^+}\text{f}(\text{x})=-\text{a}+1$
Thus, limit of f(x)exists at x=a,where a<0.
When a > 0$\lim\limits_{\text{x}\rightarrow\text{a}^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow\text{a}^-}(|\text{x}|-1)$
$=\lim\limits_{\text{x}\rightarrow\text{a}}(\text{x}-1)$ $0<\text{x}<\text{a}<\Rightarrow|\text{x}|=\text{x}$
$=\text{a}-1$
$\lim\limits_{\text{x}\rightarrow\text{a}^+}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow\text{a}^+}(|\text{x}|-1)$
$=\lim\limits_{\text{x}\rightarrow\text{a}}(\text{x}-1)$$0<\text{a}<\text{x}<\Rightarrow|\text{x}|=\text{x}$
$=\text{a}-1$
$\therefore\lim\limits_{\text{x}\rightarrow\text{a}^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow\text{a}^+}\text{f}(\text{x})=\text{a}-1$
Thus, limit of f(x) exists at x = a, where a > 0.
Thus, $\lim\limits_{\text{x}\rightarrow\text{a}}\text{f}(\text{x})$ exists for all a ≠ 0.
View full question & answer→Question 454 Marks
Evaluate the following limits in Exercise:$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}+\text{bx}}{\text{ax}+\sin\text{bx}}\text{a},\text{b},\text{a}+\text{b}\neq0,$
AnswerAt x = 0, the value of the given function takes the form $\frac{0}{0}.$ Now, $\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}+\text{bx}}{\text{ax}+\sin\text{bx}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\bigg(\frac{\sin\text{ax}}{\text{ax}}\bigg)\text{ax}+\text{bx}}{\text{ax}+\text{bx}\bigg(\frac{\sin\text{bx}}{\text{bx}}\bigg)}$ $ =\frac{\Bigg(\lim\limits_{\text{ax}\rightarrow0}\frac{\sin\text{ax}}{\text{ax}}\Bigg)\times\lim\limits_{\text{x}\rightarrow0}(\text{ax})+\lim\limits_{\text{x}\rightarrow0}\text{bx}}{\lim\limits_{\text{x}\rightarrow0}\text{ax}+{\lim\limits_{\text{x}\rightarrow0}\text{bx}\Bigg({\lim\limits_{\text{bx}\rightarrow0}\frac{\sin\text{bx}}{\text{bx}}}\Bigg)}} $ $[\text{As x}\rightarrow0\Rightarrow\text{ax}\rightarrow\text{and}\text{bx}\rightarrow0]$ $=\frac{\lim\limits_{\text{x}\rightarrow0}(\text{ax})+\lim\limits_{\text{x}\rightarrow0}\text{bx}}{\lim\limits_{\text{x}\rightarrow0}(\text{ax})+\lim\limits_{\text{x}\rightarrow0}(\text{bx})} \bigg[\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\bigg]$ $=\frac{\lim\limits_{\text{x}\rightarrow0}(\text{ax}+\text{bx})}{\lim\limits_{\text{x}\rightarrow0}(\text{ax}+\text{bx})}$ $= \lim\limits_{\text{x}\rightarrow0}(1)$ $=1$
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Find $\lim\limits_{\text{x}\rightarrow0}\text{f}(\text{x})$ and $\lim\limits_{\text{x}\rightarrow1}\text{f}(\text{x})$, where $\text{f}(\text{x})=\begin{cases}2\text{x}+3,& x \leq 0\\3(\text{x}+1), & x > 0\end{cases}$
AnswerThe given function is $\text{f}(\text{x})=\begin{cases}2\text{x}+3,& x \leq 0\\3(\text{x}+1), & x > 0\end{cases}$ $\lim\limits_{\text{x}\rightarrow0^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow0}[2\text{x}+3]=2(0)+3=3$ $\lim\limits_{\text{x}\rightarrow0^+}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow0}[3(\text{x}+1)]=3(0+1)=3$ $\therefore\lim\limits_{\text{x}\rightarrow0^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow0^+}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow0}\text{f}(\text{x})=3$ $\lim\limits_{\text{x}\rightarrow1^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow1^+}3(\text{x}+1)=3(1+1)=6$ $\lim\limits_{\text{x}\rightarrow1^+}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow1^+}3(\text{x}+1)=3(1+1)=6$ $\therefore\lim\limits_{\text{x}\rightarrow1^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow1^+}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow1}\text{f}(\text{x})=6$
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Find the derivative of $\text{x}^\text{n}+\text{ax}^\text{n-1}+\text{a}^2\text{x}^\text{n-2}+...+\text{a}^\text{n-1}\text{x}+\text{a}^\text{n}$ for some fixed real number a.
AnswerLet $\text{f}(\text{x})=\text{x}^\text{n}+\text{ax}^\text{n-1}+\text{a}^2\text{x}^\text{n-2}+...+\text{a}^\text{n-1}\text{x}+\text{a}^\text{n}$ $\therefore\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}\big(\text{x}^\text{n}+\text{ax}^\text{n-1}+\text{a}^2\text{x}^\text{n-2}+...+\text{a}^\text{n-1}\text{x}+\text{a}^\text{n}\big)$ $=\frac{\text{d}}{\text{dx}}(\text{x}^\text{n})+\text{a}\frac{\text{d}}{\text{dx}}(\text{x}^\text{n-1})+\text{a}^2\frac{\text{d}}{\text{dx}}(\text{x}^\text{n-2})+...+\text{a}^\text{n-1}\frac{\text{d}}{\text{dx}}(\text{x})+\text{a}^\text{n}\frac{\text{d}}{\text{dx}}(1)$ On using theorem $\frac{\text{d}}{\text{dx}}(\text{x})^\text{n}=\text{nx}^{\text{n}-1}$, we obtain $\text{f}'(\text{x})=\text{n}\text{x}^\text{n-1}+\text{a}(\text{n-1})\text{x}^\text{n-2}+\text{a}^2(\text{n}-2)\text{x}^\text{n-3}+...+\text{a}^\text{n-1}+\text{a}^\text{n}(0)$ $=\text{n}\text{x}^\text{n-1}+\text{a}(\text{n-1})\text{x}^\text{n-2}+\text{a}^2(\text{n}-2)\text{x}^\text{n-3}+...+\text{a}^\text{n-1}$
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Evaluate the following limits in Exercise:$\lim\limits_{\text{x}\rightarrow0}\frac{\cos2\text{x}-1}{\cos\text{x}-1}$
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{\cos2\text{x}-1}{\cos\text{x}-1}$At x = 0, the value of the given function takes the form${\frac{0}{0}}$.
$\text{Now},$$\lim\limits_{\text{x}\rightarrow0}\frac{\cos2\text{x}-1}{\cos\text{x}-1}=\lim\limits_{\text{x}\rightarrow0}\frac{1-2\sin^2\text{x}-1}{1-2\sin^2\frac{\text{x}}{2}-1} \Big[\cos\text{x}=1-2\text{sin}^2\frac{\text{x}}{2}\Big] $
$=\lim\limits_{\text{x}\rightarrow0}\frac{1-\sin^2\text{x}}{1-\sin^2\frac{\text{x}}{2}}=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\frac{\sin^2\text{x}}{\text{x}^2}\big)\times{\text{x}^2}}{\Bigg(\frac{\sin^2\frac{\text{x}}{2}}{\big(\frac{\text{x}}{2}\big)^2}\Bigg)\times\frac{\text{x}^2}{4}}$
$=4\frac{\lim\limits_{\text{x}\rightarrow0}\big(\frac{\sin^2\text{x}}{\text{x}^2}\big)}{\lim\limits_{\text{x}\rightarrow0}\Bigg(\frac{\sin^2\frac{\text{x}^2}{2}}{\big(\frac{\text{x}}{2}\big)^2}\Bigg)} $
$=4\frac{\bigg(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\bigg)^2}{\Bigg(\lim\limits_{\frac{\text{x}}{2}\rightarrow0}\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}\Bigg)^2} \big[\text{x}\rightarrow0\Rightarrow\frac{\text{x}}{2}\rightarrow0\big] $ $=4\frac{1^2}{1^2} \big[\lim\limits_{\text{y}\rightarrow0}\frac{\sin\text{y}}{\text{y}}=1\big]$$=4$
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Find the derivative of the following functions from first principle: $\cos(\text{x}-\frac{\pi}{8})$
AnswerLet $\text{f(x)}=\cos\Big(\text{x}-\frac{\pi}{8}\Big)$. Accordingly, $\text{f(x+h)}=\cos\Big(\text{x+h}-\frac{\pi}8{}\Big)$ By first principle, $\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\cos\Big(\text{x+h}-\frac{\pi}{8}\Big)-\cos\Big(\text{x}-\frac{\pi}{8}\Big)\Big]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[-2\sin\frac{\Big(\text{x+h}-\frac{\pi}{8}+\text{x}-\frac{\pi}{8}\Big)}{2}\sin\Bigg(\frac{\text{x+h}-\frac{\pi}{8}-\text{x}-\frac{\pi}{8}}{2}\Bigg)\Bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\bigg[-2\sin\bigg(\frac{\text{2x+h}-\frac{\pi}{4}}{2}\bigg)\sin\frac{\text{h}}{2}\bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\Bigg[-\sin\Bigg(\frac{2\text{x+4}-\frac{\pi}{4}}{2}\Bigg)\frac{\sin\Big(\frac{\text{h}}{2}\Big)}{\Big(\frac{\text{h}}{2}\Big)}\Bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\bigg[-\sin\bigg(\frac{\text{2x+h}-\frac{\pi}{4}}{2}\bigg)\bigg].\lim\limits_{\frac{\text{h}}{2}\rightarrow0}\frac{\sin\Big(\frac{\text{h}}{2}\Big)}{\Big(\frac{\text{h}}{2}\Big)}\ $$\Big[\text{As h}\rightarrow0\Rightarrow\frac{\text{h}}{2}\rightarrow0\Big]$ $=\sin\Big(\frac{\text{2x+0}-\frac{\pi}{4}}{2}\Big).1$ $=\sin\Big(\text{x}-\frac{\pi}{8}\Big)$
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Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\text{(ax+b)}^{\text{n}}$
AnswerLet $\text{f(x)}=\text{(ax+b)}^{\text{n}}.$ Accordingly, $\text{f(x+h)}=\big\{\text{a(x+h)+b}\big\}''=(\text{ax+ah+b)}''$ By first principle, $\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{ax+ah+b})^{\text{n}}-(\text{ax+b})^{\text{n}}}{\text{n}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{ax+b})^{\text{n}}\Big(1+\frac{\text{ah}}{\text{ax+h}}\Big)^{\text{n}}-(\text{ax+b})^{\text{n}}}{\text{h}}$ $=\text{(ax+b})^{\text{n}}\lim\limits_{\text{h}\rightarrow0}\frac{\Big(1+\frac{\text{ah}}{\text{ax+h}}\Big)^{\text{n}}-1}{\text{h}}$ $=\text{(ax+b})^{\text{n}}\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{n}}\Big[\Big\{1+\text{n}\Big(\frac{\text{ah}}{\text{ax+b}}\Big)+\frac{\text{n(n}-1)}{2!}\Big(\frac{\text{ah}}{\text{ax+b}}\Big)^2+...\Big\}-1\Big]$ (Using binomial theorem) $=\text{(ax+b})^{\text{n}}\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\text{n}\Big(\frac{\text{ah}}{\text{ax+h}}\Big)+\frac{\text{n(n}-1)\text{a}^2\text{h}^2}{2!(\text{ax+b})^2}+...$(Terms containing higher degrees of h)$\Big]$ $=\text{(ax+b})^{\text{n}}\lim\limits_{\text{h}\rightarrow0}\Big[\frac{\text{na}}{\text{(ax+b)}}+\frac{\text{n(n}-1)\text{a}^2\text{h}}{2!(\text{ax+b})^2}+...\Big]$ $=\text{(ax+b})^{\text{n}}\Big[\frac{\text{na}}{\text{(ax+b)}}+0\Big]$ $=\text{na}\frac{\text{(ax+b)}^{\text{n}}}{\text{(ax+b)}}$ $=\text{na}(\text{ax+b})^{\text{n}-1}$
View full question & answer→Question 514 Marks
Find the derivative of cos x from first principle.
AnswerLet $ \text{f}(\text{x})=\cos\text{x}$. accordingly, from the first principle, $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\cos(\text{x}+\text{h})-\cos\text{x}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\bigg[\frac{\cos\text{x}\cos\text{h}-\sin\text{x}\sin\text{h}-\cos\text{x}}{\text{h}}\bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\bigg[\frac{-\cos\text{x}(1-\cos\text{h})-\sin\text{x}\sin\text{h}}{\text{h}}\bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\bigg[\frac{-\cos\text{x}(1-\cos\text{h})}{\text{h}}-\frac{\sin\text{x}\sin\text{h}}{\text{h}}\bigg]$ $=-\cos\text{x}\bigg(\lim\limits_{\text{h}\rightarrow0}\frac{1-\cos\text{h}}{\text{h}}\bigg)-\sin\text{x}\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\sin\text{h}}{\text{h}}\bigg)$ $=-\cos\text{x}(0)-\sin\text{x}(1)$ $\bigg[\lim\limits_{\text{h}\rightarrow0}\frac{1-\cos\text{h}}{\text{h}}=0\ \text{and} \ \lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}=1\bigg]$ $=-\sin\text{x}$ $\therefore\text{f}'(\text{x})=-\sin\text{x}$
View full question & answer→Question 524 Marks
Find the derivative of the following functions: $\sec\text{x}$
Answer$\text{f}(\text{x})=\sec\text{x}$. Accordingly, from the first principle, $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\sec(\text{x}+\text{h})-\sec\text{x}}{\text{h}}$$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{1}{\cos(\text{x}+\text{h})}-\frac{1}{\cos\text{x}}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{\cos\text{x}-\cos(\text{x}+\text{h})}{\cos\text{x}\cos(\text{x}+\text{h})}\bigg]$ $=\frac{1}{\cos\text{x}}.\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{-2\sin\big(\frac{\text{x}+\text{x}+\text{h}}{2}\big)\sin\big(\frac{\text{x}-\text{x}-\text{h}}{2}\big)}{\cos(\text{x}+\text{h})}\bigg]$ $=\frac{1}{\cos\text{x}}.\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{-2\sin\big(\frac{2\text{x}+\text{h}}{2}\big)\sin\big(-\frac{\text{h}}{2}\big)}{\cos(\text{x}+\text{h})}\bigg]$ $=\frac{1}{\cos\text{x}}.\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\frac{\Bigg[-\sin\big(\frac{2\text{x}+\text{h}}{2}\big)\frac{\sin\big(\frac{\text{h}}{2}\big)}{\big(\frac{\text{h}}{2}\big)}\Bigg]}{\cos(\text{x}+\text{h})}$ $=\frac{1}{\cos\text{x}}.\lim\limits_{\frac{\text{h}}{{2}}\rightarrow0}\frac{\sin\bigg(\frac{\text{h}}{2}\bigg)}{\bigg(\frac{\text{h}}{2}\bigg)}.\lim\limits_{\text{h}\rightarrow0}\frac{\sin\bigg(\frac{2\text{x}+\text{h}}{2}\bigg)}{\cos(\text{x}+\text{h})}$ $=\frac{1}{\cos\text{x}}.1.\frac{\sin\text{x}}{\cos\text{x}}$ $=\sec\text{x}\tan\text{x}$
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For the function$\text{f}(\text{x})=\frac{\text{x}^{100}}{100}+\frac{\text{x}^{99}}{99}+...+\frac{\text{x}^2}{2}+\text{x}+1.$
Prove that $\text{f}'(1)=100$$\text{f}'(0)$.
AnswerThe given function is $\text{f}(\text{x})=\frac{\text{x}^{100}}{100}+\frac{\text{x}^{99}}{99}+...+\frac{\text{x}^2}{2}+\text{x}+1$ $\frac{\text{d}}{\text{dx}}\text{f}(\text{x})=\frac{\text{d}}{\text{dx}}\bigg[\frac{\text{x}^{100}}{100}+\frac{\text{x}^{99}}{99}+...+\frac{\text{x}^2}{2}+\text{x}+1\bigg]$ $\frac{\text{d}}{\text{dx}}\text{f}(\text{x})=\frac{\text{d}}{\text{dx}}\bigg(\frac{\text{x}^{100}}{100}\bigg)+\frac{\text{d}}{\text{dx}}\bigg(\frac{\text{x}^{99}}{99}\bigg)+...+\frac{\text{d}}{\text{dx}}\bigg(\frac{\text{x}^2}{2}\bigg)+\frac{\text{d}}{\text{dx}}\bigg(\text{x}\bigg)+\frac{\text{d}}{\text{dx}}\bigg(1\bigg)$ On using theorem $\frac{\text{d}}{\text{dx}}(\text{x})^\text{n}=\text{nx}^{\text{n}-1}$, we obtain $\frac{\text{d}}{\text{dx}}\text{f}(\text{x})=\frac{100\text{x}^{99}}{100}+\frac{99\text{x}^{98}}{99}+...+\frac{2\text{x}}{2}+1+0$ $=\text{x}^{99}+\text{x}^{98}+...+\text{x}+1$ $\therefore\text{f}'(\text{x})=\text{x}^{99}+\text{x}^{98}+...+\text{x}+1$ At $\text{x}=0$, $\text{f}'(0)=1$ At $\text{x}=1$,$\text{f}'(1)=1^{99}+1^{98}+...+1+1=[1+1+...+1+1]_{100\text{terms}}=1\times100=100$ Thus,$\text{f}'(1)=100\times\text{f}'(0)$
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Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $(\text{x}+\cos\text{x})(\text{x}-\tan\text{x})$
AnswerLet $\text{f(x)}=(\text{x}+\cos\text{x})(\text{x}-\tan\text{x})$ By product rule, $\text{f}'\text{(x)}=(\text{x}+\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{x}-\tan\text{x})+(\text{x}-\tan\text{x})\frac{\text{d}}{\text{dx}}(\text{x}+\cos\text{x})$ $=(\text{x}+\cos\text{x})\Big[\frac{\text{d}}{\text{dx}}(\text{x})-\frac{\text{d}}{\text{dx}}(\tan\text{x})\Big]+(\text{x}-\tan\text{x})(1-\sin\text{x})$ $=(\text{x}+\cos\text{x})\Big[1-\frac{\text{d}}{\text{dx}}\tan\text{x}\Big]+(\text{x}-\tan\text{x})(1-\sin\text{x})$ Let $\text{g(x)}=\tan\text{x.}$ Accordingly, $\text{g(x+h)}=\tan(\text{x+h})$ By first principle,$\text{g}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{g(x+h)}-\text{g(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\tan\text{(x+h)}-\tan\text{(x)}}{\text{h}}\Big)$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{\sin\text{(x+h)}}{\cos\text{x+h}}-\frac{\sin\text{x}}{\cos\text{x}}\Big]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{\sin\text{(x+h)}\cos\text{x}-\sin\text{x}\cos(\text{x+h})}{\cos(\text{x+h})\cos\text{x}}\Big]$ $=\frac{1}{\cos\text{x}}.\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{\sin\text{(x+h}-\text{x})}{\cos(\text{x+h})}\Big]$ $=\frac{1}{\cos\text{x}}.\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{\sin\text{(h})}{\cos(\text{x+h})}\Big]$ $=\frac{1}{\cos\text{x}}.\bigg(\lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{(h})}{\text{h}}\bigg).\Bigg(\lim\limits_{\text{h}\rightarrow0}\frac{1}{\cos(\text{x+h})}\Bigg)$ $=\frac{1}{\cos\text{x}}.1.\frac{1}{\cos(\text{x+0})}$ $=\frac{1}{\cos^2\text{x}}$ $=\sec^2\text{x}\ ...(\text{ii})$ Therefore, from (i) and (ii), we obtain $\text{f}'\text{(x)}=(\text{x}+\cos\text{x})(1-\sec^2\text{x})+(\text{x}-\tan\text{x})(1-\sin\text{x})$ $=(\text{x}+\cos\text{x})(-\tan^2\text{x})+(\text{x}-\tan\text{x})(1-\sin\text{x})$ $=-\tan^2\text{x}(\text{x}+\cos\text{x})+(\text{x}-\tan\text{x})(1-\sin\text{x})$
View full question & answer→Question 554 Marks
Find the derivative of the following functions:$\text{f}(\text{x}) =5\sec\text{x}+4\cos\text{x}$
AnswerLet $\text{f}(\text{x}) =5\sec\text{x}+4\cos\text{x}$. Accordingly, from the first principle, $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{5\sec(\text{x}+\text{h})+4\cos(\text{x}+\text{h})-[5\sec\text{x}+4\cos\text{x}]}{\text{h}}$ $=5\lim\limits_{\text{h}\rightarrow0}\frac{[\sec(\text{x}+\text{h})-\sec\text{x}]}{\text{h}}+4\lim\limits_{\text{h}\rightarrow0}\frac{[\cos(\text{x}+\text{h})-\cos\text{x}]}{\text{h}}$ $=5\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{1}{\cos(\text{x}+\text{h})}-\frac{1}{\cos\text{x}}\bigg]+4\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}[\cos(\text{x}+\text{h})-\cos\text{x}]$$=\frac{5}{\cos\text{x}}\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{\cos\text{x}-\cos(\text{x}+\text{h})}{\cos\text{x}\cos(\text{x}+\text{h})}\bigg]+4\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}[\cos{\text{x}}\cos\text{h}-\sin\text{x}\sin\text{h}-\cos\text{x}] $
$=\frac{5}{\cos\text{x}}\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{-2\sin\bigg(\frac{\text{x}+\text{x}+\text{h}}{2}\bigg)\sin\bigg (\frac{\text{x}-\text{x}-\text{h}}{2}\bigg)}{\cos(\text{x}+\text{h})}\bigg]+4\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}[-\cos{\text{x}}(1-\cos\text{h})-\sin\text{x}\sin\text{h}] $$=\frac{5}{\cos\text{x}}\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{-2\sin\bigg(\frac{2\text{x}+\text{h}}{2}\bigg)\sin\bigg (-\frac{\text{h}}{2}\bigg)}{\cos(\text{x}+\text{h})}\bigg]+4\bigg[-\cos\lim\limits_{{\text{h}}\rightarrow0}\frac{(1-\cos\text{h})}{\text{h}}-\sin\text{x}\lim\limits_{{\text{h}}\rightarrow0}\frac{\sin\text{h}}{\text{h}}\bigg] $ $=\frac{5}{\cos\text{x}}\lim\limits_{{\text{h}}\rightarrow0}\Bigg[\frac{\sin\bigg(\frac{2\text{x}+\text{h}}{2}\bigg).\frac{\sin\bigg (\frac{\text{h}}{2}\bigg)}{\frac{\text{h}}{2}}}{\cos(\text{x}+\text{h})}\Bigg]+4[(-\cos{\text{x}).(0)}-(\sin\text{x}).1] $ $=\frac{5}{\cos\text{x}}.\Bigg[\lim\limits_{\text{h}\rightarrow0}\frac{\sin\frac{2\text{x}+\text{h}}{2}}{\cos(\text{x}+\text{h})}.\lim\limits_{\text{h}\rightarrow0}\frac{\sin\big(\frac{\text{h}}{2}\big)}{\frac{\text{h}}{2}}\Bigg]-4\sin\text{x}$ $=\frac{5}{\cos\text{x}}.\frac{\sin\text{x}}{\cos\text{x}}.1-4\sin\text{x}$ $=5\sec\text{x}\tan\text{x.}-4\sin\text{x}$
View full question & answer→Question 564 Marks
Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\frac{\sin{(\text{x+a)}}}{\cos\text{x}}$
AnswerLet $\text{f(x)}=\frac{\sin{(\text{x+a})}}{\cos\text{x}}$ By quotient rule, $\text{f}'\text{(x)}=\frac{\cos\text{x}\frac{\text{d}}{\text{dx}}\big[\sin(\text{x+a)}\big]-\sin\text{(x+a)}\frac{\text{d}}{\text{dx}}\cos\text{x}}{\cos^2\text{x}}$ $\text{f}'\text{(x)}=\frac{\cos\text{x}\frac{\text{d}}{\text{dx}}\big[\sin(\text{x+a)}\big]-\sin\text{(x+a)}(-\sin\text{x})}{\cos^2\text{x}}\ ...(\text{i})$ $\text{Let g(x)}=\sin\text{(x+a). Accordingly, (x+h)}=\sin(\text{x+h+a})$ By first principle, $\text{g}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{g(x+h)}-\text{g(x)}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\big[\sin(\text{x+h+a})-\sin(\text{x+a})\big]$$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[2\cos\Big(\frac{\text{x+h+a+x+a}}{2}\Big)\sin\Big(\frac{\text{x+h+a}-\text{x}-\text{a}}{2}\Big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[2\cos\Big(\frac{\text{2x+2a+h}}{2}\Big)\sin\Big(\frac{\text{h}}{2}\Big)\Big]$$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\cos\Big(\frac{\text{2x+2a+h}}{2}\Big)\sin\Bigg\{\frac{\Big(\sin\frac{\text{h}}{2}\Big)}{\Big(\frac{\text{h}}{2}\Big)}\Bigg\}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\cos\Big(\frac{\text{2x+2a+h}}{2}\Big)\lim\limits_{\frac{\text{h}}{2}\rightarrow0}\Bigg\{\frac{\Big(\sin\frac{\text{h}}{2}\Big)}{\Big(\frac{\text{h}}{2}\Big)}\Bigg\}$ $\Big[\text{As h}\rightarrow0\Rightarrow\frac{\text{h}}{2}\rightarrow0\Big]$ $=\Big(\cos\frac{\text{2x+2a}}{2}\Big)\times1$ $\Big[\lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}=1\Big]$$=\cos(\text{x+a})\ ...(\text{ii})$
From (i) and (ii), we obtain $\text{f}'\text{(x)}=\frac{\cos\text{x}.\cos(\text{x+a})+\sin\text{x}\sin(\text{x+a})}{\cos^2\text{x}}$ $=\frac{\cos(\text{x+a}-\text{x})}{\cos^2\text{x}}$ $=\frac{\cos\text{a}}{\cos^2\text{x}}$
View full question & answer→Question 574 Marks
Find the derivative of the following functions: $\text{cosecx}$
AnswerLet $\text{f}(\text{x}) = 5\sec\text{x}+4\cos\text{x}$. Accordingly, from the first principle, $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$ $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}[\text{cosec}(\text{x}+\text{h})-\text{cosec}\text{x}]$ $=\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{1}{\sin(\text{x}+\text{h})}-\frac{1}{\sin\text{x}}\bigg]$ $=\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{\sin\text{x}-\sin(\text{x}+\text{h})}{\sin(\text{x}+\text{h})\sin\text{x}}\bigg]$ $=\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{2\cos\bigg(\frac{\text{x}+\text{x}+\text{h}}{2}\bigg)\sin\bigg (\frac{\text{x}-\text{x}-\text{h}}{2}\bigg)}{\sin(\text{x}+\text{h})\sin\text{x}}\bigg]$ $=\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{2\cos\bigg(\frac{2\text{x}+\text{h}}{2}\bigg)\sin\bigg (-\frac{\text{h}}{2}\bigg)}{\sin(\text{x}+\text{h})\sin\text{x}}\bigg]$ $=\lim\limits_{{\text{h}}\rightarrow0}\Bigg[\frac{-\cos\bigg(\frac{2\text{x}+\text{h}}{2}\bigg).\frac{\sin\bigg (\frac{\text{h}}{2}\bigg)}{\frac{\text{h}}{2}}}{\sin(\text{x}+\text{h})\sin\text{x}}\Bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{-\cos\big(\frac{2\text{x}+\text{h}}{2}\big)}{\sin(\text{x}+\text{h})\sin\text {x}}\Bigg).\lim\limits_{\text{h}\rightarrow0}\frac{\sin\big(\frac{\text{h}}{2}\big)}{\big(\frac{\text{h}}{2}\big)}$ $=\bigg(\frac{-\cos\text{x}}{\sin\text{x}\sin\text{x}}\bigg).1$
View full question & answer→Question 584 Marks
Find the derivative of the following functions: $ 2 \tan\text{x} – 7\sec\text{x}$
AnswerLet $\text{f}(\text{x})= 2 \tan\text{x} – 7\sec\text{x}$ Accordingly, from the first principle, $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\big[2\tan(\text{x}+\text{h})-7\sec(\text{x}+\text{h})-2\tan\text{x}+7\sec\text{x}\big]$$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\big[2\big\{\tan(\text{x}+\text{h})-\tan\text{x}\big\}-7\big\{\sec(\text{x}+\text{h})-\sec\text{x}\big\}\big]$
$=2\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\big[\big\{\tan(\text{x}+\text{h})-\tan\text{x}\big\}-7\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\big\{\sec(\text{x}+\text{h})-\sec\text{x}\big\}\big]$
$=2\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{\sin(\text{x}+\text{h})}{\cos(\text{x}+\text{h})}-\frac{\sin\text{x}}{\cos\text{x}}\bigg]-7\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{1}{\cos(\text{x}+\text{h})}-\frac{1}{\cos\text{x}}\Bigg]$
$$$=2\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{\sin(\text{x}+\text{h})\cos\text{x}-\sin\text{x}\cos(\text{x}+\text{h})}{\cos\text{x}\cos(\text{x}+\text{h})}\bigg]-7\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{{\cos\text{x}}-{\cos(\text{x}+\text{h})}}{{\cos\text{x}}\cos(\text{x}+\text{h})}\Bigg]$
$=2\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{\sin(\text{x}+\text{h}-\text{x})}{\cos\text{x}\cos(\text{x}+\text{h})}\bigg]-7\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{-2\sin\bigg(\frac{\text{x}+\text{x}+\text{h}}{2}\bigg)+\sin\bigg(\frac{\text{x}+\text{x}-\text{h}}{2}\bigg)}{{\cos\text{x}}\cos(\text{x}+\text{h})}\Bigg]$ $=2\lim\limits_{\text{h}\rightarrow0}\bigg[\bigg(\frac{\sin\text{x}}{\text{h}}\bigg)\frac{1}{\cos\text{x}\cos(\text{x}+\text{h})}\bigg]-7\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{-2\sin\bigg(\frac{2\text{x}+\text{h}}{2}\bigg)+\sin\bigg(-\frac{\text{h}}{2}\bigg)}{{\cos\text{x}}\cos(\text{x}+\text{h})}\Bigg]$$=2\bigg(\lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{x}}{\text{h}}\bigg)\bigg(\lim\limits_{\text{h}\rightarrow0}\frac{1}{\cos\text{x}\cos(\text{x}+\text{h})}\bigg)-7\Bigg(\lim\limits_{\frac{\text{h}}{2}\rightarrow0}\frac{\sin\frac{\text{h}}{\text{h}}{2}}{2}\Bigg)\Bigg(\lim\limits_{\text{h}\rightarrow0}\frac{\sin\bigg(\frac{2\text{x}+\text{h}}{2}\bigg)}{{\cos\text{x}}\cos(\text{x}+\text{h})}\Bigg)$
$=2\bigg(\lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{x}}{\text{h}}\bigg)\bigg(\lim\limits_{\text{h}\rightarrow0}\frac{1}{\cos\text{x}\cos(\text{x}+\text{h})}\bigg)-7\Bigg(\lim\limits_{\frac{\text{h}}{2}\rightarrow0}\frac{\sin\frac{\text{h}}{2}}{\frac{\text{h}}{2}}\Bigg)\Bigg(\lim\limits_{\text{h}\rightarrow0}\frac{\sin\bigg(\frac{2\text{x}+\text{h}}{2}\bigg)}{{\cos\text{x}}\cos(\text{x}+\text{h})}\Bigg)$$=2.1.\frac{1}{\cos\text{x}\cos\text{x}}-7.1\bigg(\frac{\sin\text{x}}{\cos\text{x}\cos\text{x}}\bigg)$
$=2\sec^2\text{x}-7\sec\text{x}\tan\text{x}$
View full question & answer→Question 594 Marks
Find the derivative of the following functions: $5\sin\text{x} – 6\cos\text{x} + 7$
AnswerLet $\text{f}(\text{x})=5\sin\text{x} – 6\cos\text{x} + 7$ Accordingly, from the first principle, $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\big[5\sin(\text{x}+\text{h})-6\cos(\text{x}+\text{h})+7-5\sin\text{x}+6\cos\text{x}-7\big]$$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\big[5\big\{\sin(\text{x}+\text{h})-\sin\text{x}\big\}-6\big\{\cos(\text{x}+\text{h})-\cos\text{x}\big\}\big]$
$=5\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\big[\big\{\sin(\text{x}+\text{h})-\sin\text{x}\big\}-6\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\big\{\cos(\text{x}+\text{h})-\cos\text{x}\big\}\big]$
$=5\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\bigg[2\cos\bigg(\frac{\text{x}+\text{h}+\text{x}}{2}\bigg)\sin\bigg(\frac{\text{x}+\text{h}-\text{x}}{2}\bigg)\bigg]-6\lim\limits_{\text{h}\rightarrow0}\frac{\cos\text{x}\cos\text{h}-\sin\text{x}\sin\text{h}-\cos\text{x}}{\text{h}}$
$$$=5\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\bigg[2\cos\bigg(\frac{2\text{x}+\text{h}}{2}\bigg)\sin\frac{\text{h}}{2}\bigg]-6\lim\limits_{\text{h}\rightarrow0}\Bigg[\frac{-\cos\text{x}(1-\cos\text{h})-\sin\text{x}\sin\text{h}}{\text{h}}\Bigg]$
$=5\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\bigg[2\cos\bigg(\frac{2\text{x}+\text{h}}{2}\bigg)\frac{\sin\frac{\text{h}}{2}}{\frac{\text{h}}{2}}\bigg]-6\lim\limits_{\text{h}\rightarrow0}\Bigg[\frac{-\cos\text{x}(1-\cos\text{h})}{{\text{h}}}-{\frac{-\sin\text{x}\sin\text{h}}{\text{h}}}\Bigg]$ $=5\lim\limits_{\text{h}\rightarrow0}\bigg[\cos\bigg(\frac{2\text{x}+\text{h}}{2}\bigg)\bigg]\Bigg[\lim\limits_{\frac{\text{h}}{2}\rightarrow0}\frac{\sin\frac{\text{h}}{2}}{\frac{\text{h}}{2}}\Bigg]-6\Bigg[(-\cos\text {x})\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{1-\cos\text{h}}{{\text{h}}}\bigg)-\sin\text{x}\lim\limits_{\text{h}\rightarrow0}{\bigg(\frac{\sin\text{h}}{\text{h}}\bigg)}\Bigg]$ $=5\cos\text{x}.1-6\big[(-\cos\text{x}).(0)-\sin\text{x}.1\big]$ $=5\cos\text{x}+6\sin\text{x}$
View full question & answer→Question 604 Marks
Find the derivative of the following functions: $3\cot\text{x} + 5\text{cosec}\text{x}$
AnswerLet $\text{f}(\text{x})=3\cot\text{x} + 5\text{cosec}\text{x}$. Accordingly, from the first principle, $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{3\cot(\text{x}+\text{h})+5\text{cosec}(\text{x}+\text{h})-3\cot\text{x}-5\text{cosec}\text{x}}{\text{h}}$ $=3\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}[{\cot(\text{x}+\text{h})-\cot\text{x}]+5\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}[\text{cosec}(\text{x}+\text{h})-\text{cosec}\text{x}}]$...(1) Now, $\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}[{\cot(\text{x}+\text{h})-\cot\text{x}}]$ $=\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{\cos(\text{x}+\text{h})}{\sin(\text{x}+\text{h})}-\frac{\cos\text{x}}{\sin\text{x}}\bigg]$ $=\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{\cos(\text{x}+\text{h})\sin\text{x}-\cos\text{x}\sin(\text{x}+\text{h})}{\sin\text{x}\sin(\text{x}+\text{h})}\bigg]$ $=\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{\sin(\text{x}-\text{x}-\text{h})}{\sin\text{x}\sin(\text{x}+\text{h})}\bigg]$$=\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{\sin(-\text{h})}{\sin\text{x}\sin(\text{x}+\text{h})}\bigg]$
$=\Bigg(-\lim\limits_{{\text{h}}\rightarrow0}\frac{\sin\text{h}}{\text{h}}\Bigg).\Bigg(\lim\limits_{{\text{h}}\rightarrow0}(\frac{1}{\sin\text{x}.\sin(\text{x}+\text{h})}\Bigg)$
$=-1.\frac{1}{\sin\text{x}.\sin(\text{x}+0)}=\frac{-1}{\sin^2\text{x}}=-\text{cosec}^2\text{x}$ ...(2)
$\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}[\text{cosec}(\text{x}+\text{h})-\text{cosecx}]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{1}{\sin(\text{x}+\text{h})}-\frac{1}{\sin\text{x}}\bigg]$ $=\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{\sin\text{x}-\sin(\text{x}-\text{h})}{\sin(\text{x}+\text{h})\sin\text{x}}\bigg]$ $=\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{2\cos\bigg(\frac{\text{x}+\text{x}+\text{h}}{2}\bigg).\sin\frac{(\text{x}-\text{x}-\text{h})}{2}}{\sin(\text{x}+\text{h})\sin\text{x}}\bigg]$ $=\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{2\cos\bigg(\frac{2\text{x}+\text{h}}{2}\bigg).\sin\bigg(-\frac{\text{h}}{2}\bigg)}{\sin(\text{x}+\text{h})\sin\text{x}}\bigg]$ $=\lim\limits_{{\text{h}}\rightarrow0}{\text{h}}\frac{-\cos\bigg(\frac{2\text{x}+\text{h}}{2}\bigg).\frac{\sin\bigg(\frac{\text{h}}{2}\bigg)}{\bigg(\frac{\text{h}}{2}\bigg)}}{\sin(\text{x}+\text{h})\sin\text{x}}$ $=\lim\limits_{{\text{h}}\rightarrow0}\Bigg(\frac{-\cos\bigg(\frac{2\text{x}+\text{h}}{2}\bigg)}{\sin(\text{x}+\text{h})\sin\text{x}}\Bigg).\lim\limits_{\frac{{\text{h}}}{2}\rightarrow0}\frac{\sin\bigg(\frac{\text{h}}{2}\bigg)}{\bigg(\frac{\text{h}}{2}\bigg)}$$=\bigg(\frac{-\cos\text{x}}{\sin\text{x}\sin\text{x}}\bigg).1$
$=-\text{cosecx}\cot\text{x}$ ...(3)
From (1), (2), and (3), we obtain $\text{f}'(\text{x})=-3\text{cosec}^2-5\text{cosecx}\cot\text{x}$
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