Question 12 Marks
Differentiate the following function with respect to $(\text{x})$:$\log_3\text{x}+3\log_\text{e}\text{x}+2\tan\text{x}$
AnswerWe have,$\frac{\text{d}}{\text{dx}}(\log_3\text{x}+3\log_\text{e}\text{x}+2\tan\text{x})$
$=\frac{1}{\log3}\frac{\text{d}}{\text{dx}}(\log\text{x})+3\frac{\text{d}}{\text{dx}}(\log_\text{e}\text{x})+2\frac{\text{d}}{\text{dx}}(\tan\text{x})\ \Bigg[\because\log_3\text{x}=\frac{\log\text{x}}{\log3}\Bigg]$
$=\frac{1}{\log3}\times\frac{1}{\text{x}}+\frac{3}{\text{x}}+2\sec^2\text{x}$
$=\frac{1}{\text{x}\log3}+\frac{3}{\text{x}}+2\sec^2\text{x}$
View full question & answer→Question 22 Marks
Differentiate the following function with respect to $\text{x}$:$\text{x}^3\text{e}^\text{x}$
AnswerLet $\text{u}=\text{x}^3;\text{v}=\text{e}^\text{x}$Then, $\text{u}'=3\text{x}^2;\text{v}'=\text{e}^\text{x}$
Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$
$\frac{\text{d}}{\text{dx}}(\text{x}^3\text{e}^\text{x})=\text{x}^3\text{e}^\text{x}+\text{e}^\text{x}(3\text{x}^2)$
$=\text{x}^2\text{e}^\text{x}(\text{x}+3)$
View full question & answer→Question 32 Marks
Differentiate the following functions with respect to x:$\frac{\text{e}^\text{x}}{1+\text{x}^2}$
Answer$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{e}^\text{x}}{1+\text{x}^2}\Big)$$=\frac{(1+\text{x}^2)\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})-(\text{e}^\text{x})\frac{\text{d}}{\text{dx}}(1+\text{x}^2)}{(1+\text{x}^2)^2}$
$=\frac{(1+\text{x}^2)\text{e}^\text{x}-\text{e}^\text{x}\times\text{2x}}{(1+\text{x}^2)^2}$
$=\frac{\text{e}^\text{x}(1+\text{x}^2-\text{2x})}{(1+\text{x}^2)^2}$
$=\frac{\text{e}^\text{x}(1-\text{x})^2}{(1+\text{x}^2)^2}$
View full question & answer→Question 42 Marks
Differentiate the following function with respect to $(\text{x})$:$\frac{1}{\sin\text{x}}+2^{\text{x}+3}+\frac{4}{\log\text{x}^3}$
AnswerWe have,$\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\sin\text{x}}+2^{\text{x}+3}+\frac{4}{\log\text{x}^3}\Big)$
$=\frac{\text{d}}{\text{dx}}\text{cosec}\text{x}+2^2\frac{\text{d}}{\text{dx}}(2^{\text{x}})+\frac{4}{\log3}\times\frac{\text{d}}{\text{dx}}(\log\text{x})\ \Big[\because\log_\text{b}\text{a}=\frac{\log\text{a}}{\log\text{b}}\Big]$
$=-\text{coesc}\text{x}.\cot\text{x}+8.2^\text{x}\log2+\frac{4}{\log3}\times\frac{1}{\text{x}}\ \Big[\because\frac{\text{d}}{\text{dx}}(\text{a}^\text{x})=\text{a}^\text{x}\log\text{a}\Big]$
$=-\text{coesc}\text{x}.\cot\text{x}+2^{\text{x}+3}\log2+\frac{4}{\text{x}\log3}$
View full question & answer→Question 52 Marks
Differentiate the following function with respect to $(\text{x})$:$2\sec\text{x}+3\cot\text{x}-4\tan\text{x}$
AnswerWe have,$\frac{\text{d}}{\text{dx}}(2\sec\text{x}+3\cot\text{x}-4\tan\text{x})$
$=2\frac{\text{d}}{\text{dx}}(\sec\text{x})+3\frac{\text{d}}{\text{dx}}(\cot\text{x})-4\frac{\text{d}}{\text{dx}}(\tan\text{x})$
$=2\sec\text{x}\tan\text{x}-3\text{cosec}^2\text{x}-4\sec^2\text{x}$
View full question & answer→Question 62 Marks
Differentiate the following functions with respect to x:$\frac{\text{x}\tan\text{x}}{\sec\text{x}+\tan\text{x}}$
AnswerWe have, $\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}\tan\text{x}}{\sec\text{x}+\tan\text{x}}\Big)$
Using Quotient rule, we get
$=\frac{(\sec\text{x}+\tan\text{x})\frac{\text{d}}{\text{dx}}(\text{x}\tan\text{x})-(\text{x}\tan\text{x})\frac{\text{d}}{\text{dx}}(\sec\text{x}+\tan\text{x}){}}{(\sec\text{x}+\tan\text{x})^2}$
$=\frac{(\sec\text{x}+\tan\text{x})(\text{x}\sec^2\text{x}+\tan\text{x})-(\text{x}\tan\text{x})(\sec\text{x}\tan\text{x}+\sec^2\text{x})}{(\sec\text{x}+\tan\text{x})^2}$ [Used product rule]
$=\frac{(\sec\text{x}+\tan\text{x})(\text{x}\sec^2\text{x}+\tan\text{x})-\text{x}\sec\text{x}+\tan^2\text{x}-\text{x}\tan\text{x}\sec^2\text{x})}{(\sec\text{x}+\tan\text{x})^2}$
$=\frac{(\sec\text{x}+\tan\text{x})(\text{x}\sec^2\text{x}+\tan\text{x})-(\text{x}\tan\text{x})(\sec\text{x}\tan\text{x}+\sec^2\text{x})}{(\sec\text{x}+\tan\text{x})^2}$
$=\frac{(\sec\text{x}+\tan\text{x})(\text{x}\sec^2\text{x}+\tan\text{x})-\text{x}\tan\text{x}\sec\text{x}(\sec\text{x}+\tan\text{x})}{(\sec\text{x}+\tan\text{x})^2}$
$=\frac{(\text{x}\sec^2\text{x}+\tan\text{x}-\text{x}\tan\text{x}\sec\text{x})(\sec\text{x}+\tan\text{x})}{(\sec\text{x}+\tan\text{x})^2}$
$=\frac{(\text{x}\sec^2\text{x}+\tan\text{x}-\text{x}\tan\text{x}\sec\text{x})}{(\sec\text{x}+\tan\text{x})}$
$=\frac{\text{x}\sec\text{x}(\sec\text{x}-\tan\text{x})+\tan\text{x}}{(\sec\text{x}+\tan\text{x})}$
View full question & answer→Question 72 Marks
Differentiate the following function with respect to $(\text{x})$:$\frac{\text{x}^3}{3}-2\sqrt{\text{x}}+\frac{5}{\text{x}^2}$
AnswerWe have to differentiate $\text{f}(\text{x})$ with respect to $\text{x}$$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^3}{3}-2\sqrt{\text{x}}+\frac{5}{\text{x}^2}\Big)$
$\frac{1}{3}\frac{\text{d}}{\text{dx}}(\text{x}^3)-2\frac{\text{d}}{\text{dx}}\sqrt{\text{x}}+5\frac{\text{d}}{\text{dx}}(\text{x}^{-2})$
$=\frac{1}{3}.3\text{x}^2-2\frac{1.1}{2\sqrt{\text{x}}}+5.(-2)\text{x}^{-3}$
$=\text{x}^2-\text{x}^{-\frac{1}{2}}-10\text{x}^{-3}$
$=\text{x}^2-\frac{1}{\sqrt{\text{x}}}-\frac{10}{\text{x}^3}$
View full question & answer→Question 82 Marks
AnswerWe have,
$\text{f(x)}=\text{x}$
$\because\text{f}'\text{(a)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(a+h)}-\text{f(a)}}{\text{h}}$
$\therefore\text{f}'\text{(1)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(1+h)}-\text{f(1)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{1+h}-1}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0} 1$
$\therefore\text{f}'(1)=1$
View full question & answer→Question 92 Marks
Differentiate the following functions with respect to x:$\frac{\text{x}+\text{e}^\text{x}}{1+\log\text{x}}$
Answer$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}+\text{e}^\text{x}}{1+\log\text{x}}\Big)$$=\frac{(1+\log\text{x})\frac{\text{d}}{\text{dx}}(\text{x}+\text{e}^\text{x})-(\text{x}+\text{e}^\text{x})\frac{\text{d}}{\text{dx}}(1+\log\text{x})}{(1+\log\text{x})^2}$
$=\frac{(1+\log\text{x})(1+\text{e}^\text{x})-(\text{x}+\text{e}^\text{x})\times\frac{1}{\text{x}}}{(1+\log\text{x})^2}$
$=\frac{\text{x}(1+\log\text{x}+\text{e}^\text{x}+\text{e}^\text{x}\log\text{x})-\text{x}-\text{e}^\text{x}}{\text{x}(1+\log\text{x})^2}$
$=\frac{\text{x}+\text{x}\log\text{x}+\text{x}\text{e}^\text{x}+\text{x}\text{e}^\text{x}\log\text{x}-\text{x}-\text{e}^\text{x}}{\text{x}(1+\log\text{x})^2}$
$=\frac{\text{x}\log\text{x}(1+\text{e}^\text{x})-\text{e}^\text{x}(1-\text{x})}{\text{x}(1+\log\text{x})^2}$
View full question & answer→Question 102 Marks
Differentiate the following function with respect to x:$\text{x}^2\sin\text{x}\log\text{x}$
AnswerLet $\text{u}=\text{x}^\text{2};\text{v}=\sin\text{x};\text{w}=\log\text{x}$Then, $\text{u}'=2\text{x};\text{v}'=\cos\text{x};\text{w}'=\frac{1}{\text{x}}$
Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uvw})=\text{u}'\text{vw}+\text{uw}\text{v}'+\text{uv}\text{w}'$
$\frac{\text{d}}{\text{dx}}(\text{x}^2\sin\text{x}\log\text{x})=2\text{x}\sin\text{x}\log\text{x}+\text{x}^2\cos\text{x}\log\text{x}+\text{x}^2\sin\text{x}.\frac{1}{\text{x}}$
$=2\text{x}\sin\text{x}\log\text{x}+\text{x}^2\cos\text{x}\log\text{x}+\text{x}\sin\text{x}$
View full question & answer→Question 112 Marks
Differentiate the following function with respect to $\text{x}:$$\text{x}^\text{n}\tan\text{x}$
AnswerLet $\text{u}=\text{x}^\text{n};\text{v}=\tan\text{x}$Then, $\text{u}'=\text{nx}^{\text{n}-1};\text{v}'=\sec^2\text{x}$
Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$
$\frac{\text{d}}{\text{dx}}(\text{x}^\text{n}\tan\text{x})=\text{x}^\text{n}\sec^2\text{x}+\tan\text{x}(\text{nx}^{\text{n}-1})$
$=\text{x}^{\text{n}-1}(\text{x}\sec^2\text{x}+\text{n}\tan\text{x})$
View full question & answer→Question 122 Marks
Differentiate the following functions with respect to x:$\frac{1}{\text{ax}^2+\text{bx}+\text{c}}$
Answer$\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\text{ax}^2+\text{bx}+\text{c}}\Big)$$=\frac{(\text{ax}^2+\text{bx}+\text{c})\frac{\text{d}}{\text{dx}}(1)-1\frac{\text{d}}{\text{dx}}(\text{ax}^2+\text{bx}+\text{c})}{(\text{ax}^2+\text{bx}+\text{c})^2}$
$=\frac{-(\text{2ax}+\text{b})}{(\text{ax}^2+\text{bx}+\text{c})^2}$
$\therefore\frac{\text{d}}{\text{dx}}\frac{1}{\text{ax}^2+\text{bx}+\text{c}}=\frac{-(\text{2ax}+\text{b})}{(\text{ax}^2+\text{bx}+\text{c})^2}$
View full question & answer→Question 132 Marks
Differentiate the following function with respect to x:$\text{x}^5\text{e}^\text{x}+\text{x}^6\log\text{x}$
Answer$\frac{\text{d}}{\text{dx}}=(\text{x}^5\text{e}^\text{x}+\text{x}^6\log\text{x})$$=\text{x}^5\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})+\text{e}^\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^5)+(\text{x})^6\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^6)$
$=\text{x}^5\text{e}^\text{x}+\text{e}^\text{x}(5\text{x}^4)+\text{x}^6.\frac{1}{\text{x}}+\log\text{x}(6\text{x}^5)$
$=\text{x}^5\text{e}^\text{x}+\text{e}^\text{x}(5\text{x}^4)+\text{x}^5+\log\text{x}(6\text{x}^5)$
$=\text{x}^4(\text{x}\text{e}^\text{x}+5\text{e}^\text{x}+\text{x}+\text{6x}+\log\text{x})$
View full question & answer→Question 142 Marks
Differentiate the following functions with respect to x:$\frac{\text{2x}-1}{\text{x}^2+1}$
Answer$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{2x}-1}{\text{x}^2+1}\Big)=\frac{(\text{x}^2+1)\frac{\text{d}}{\text{dx}}(\text{2x}-1)-(\text{2x}-1)\frac{\text{d}}{\text{dx}}(\text{x}^2+1)}{(\text{x}^2+1)^2}$$=\frac{(\text{x}^2+1)\times2-(\text{2x}-1)\times\text{2x}}{(\text{x}^2+1)^2}$
$=\frac{\text{2x}^2+2-\text{4x}^2+\text{2x}}{(\text{x}^2+1)^2}$
$=\frac{-\text{2x}+\text{2x}+2}{(\text{x}^2+1)^2}$
$=\frac{2(-\text{x}^2+\text{x}+1)}{(\text{x}^2+1)^2}$
$=\frac{2(1+\text{x}-\text{x}^2)}{(\text{x}^2+1)^2}$
View full question & answer→Question 152 Marks
Differentiate the following function with respect to $(\text{x})$:$\text{e}^{\text{x}\log\text{a}}+\text{e}^{\text{a}\log\text{x}}+{\text{e}^{\text{a}\log\text{a}}}$
AnswerWe have,$\frac{\text{d}}{\text{dx}}(\text{e}^{\text{x}\log\text{a}}+\text{e}^{\text{a}\log\text{x}}+{\text{e}^{\text{a}\log\text{a}}})$
$=\frac{\text{d}}{\text{dx}}(\text{e}^{\text{x}\log\text{a}})+\frac{\text{d}}{\text{dx}}(\text{e}^{\text{a}\log\text{x}})+\frac{\text{d}}{\text{dx}}(\text{e}^{\text{a}\log\text{a}})$
$=\text{e}^{\text{x}\log\text{a}}.\log\text{a}+\text{e}^{\text{a}\log\text{x}}.\frac{\text{a}}{\text{x}}+0\ [\because\text{e}^{\text{a}\log\text{a}}$is constant$]$
$=\log\text{a}.\text{e}^{\text{x}\log\text{a}}+\frac{\text{a}}{\text{x}}\text{e}^{\text{a}\log\text{x}}$
$=\log\text{a}.\text{a}^\text{x}+\frac{\text{a}}{\text{x}}\text{x}^\text{a}\ [\text{a}^\text{x}$can be written as $\text{e}^{\text{x}\log\text{a}}]$
$=\text{a}^\text{x}\log\text{a}+\text{ax}^{\text{a}-1}$
View full question & answer→Question 162 Marks
Differentiate the following functions with respect to x:$\frac{\text{ax}^2+\text{bx}+\text{c}}{\text{px}^2+\text{qx}+\text{r}}$
Answer$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{ax}^2+\text{bx}+\text{c}}{\text{px}^2+\text{qx}+\text{r}}\Big)$$=\frac{(\text{px}^2+\text{qx}+\text{r})\frac{\text{d}}{\text{dx}}(\text{ax}^2+\text{bx}+\text{c})-(\text{ax}^2+\text{bx}+\text{c})\frac{\text{d}}{\text{dx}}(\text{px}^2+\text{qx}+\text{r})}{(\text{px}^2+\text{qx}+\text{r})^2}$
$=\frac{(\text{px}^2+\text{qx}+\text{r})(\text{2ax}+\text{b})-(\text{ax}^2+\text{bx}+\text{c})(\text{2px}+\text{q})}{(\text{px}^2+\text{qx}+\text{r})^2}$
$=\frac{2\text{apx}^3+\text{2aqx}^2+\text{2axr}+\text{bpx}^2+\text{bqx}+\text{br}-(\text{2apx}^3+\text{2pbx}^2+\text{2pcx}+\text{qax}^2+\text{bqx}+\text{cq})}{(\text{px}^2+\text{qx}+\text{r})^2}$
$=\frac{\text{2apx}^3-\text{2apx}^3+\text{2aqx}^2+\text{bpx}^2-\text{2pbx}^2-\text{qax}^2+\text{2arx}+\text{bqx}-\text{2pcx}-\text{bqx}+\text{br}-\text{cq}}{(\text{px}^2+\text{qx}+\text{r})^2}$
$=\frac{\text{aqx}^2-\text{bpx}^2+\text{2axr}-\text{2cpx}+\text{br}-\text{cq}}{(\text{px}^2+\text{qx}+\text{r})^2}$
$\frac{\text{x}^2(\text{aq}-\text{bp})+2(\text{ar}-\text{cp})\text{x}+\text{br}-\text{cq}}{(\text{px}^2+\text{qx}+\text{r})^2}$
$\frac{(\text{aq}-\text{bp})\text{x}^2+2(\text{ar}-\text{cp})\text{x}+\text{br}-\text{cq}}{(\text{px}^2+\text{qx}+\text{r})^2}$
View full question & answer→Question 172 Marks
Find the derivative of f(x) = 3x at x = 2
AnswerWe have,
$\text{f(x)} = 3\text{x}$
$\because \text{f}'(\text{a}) = \lim\limits_{\text{h} \rightarrow0} \frac{\text{f(a+h)}-\text{f(a)}}{\text{h}}$
$\therefore \text{f}'\text{(a)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(2+h)}-\text{f(2)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{3(2\text{+h})-6}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{3h}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0} 3$
$\therefore\text{f}'{(2)}=3$
View full question & answer→Question 182 Marks
Differentiate the following function with respect to $\text{x}:$$\text{x}^\text{n}\log_\text{a}\text{x}$
AnswerLet $\text{u}=\text{x}^\text{n};\text{v}=\log_\text{a}\text{x}=\frac{\log\text{x}}{\log\text{a}}$Then, $\text{u}'=\text{nx}^{\text{n}-1};\text{v}'=\frac{1}{\text{x}\log\text{a}}$
Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$
$\frac{\text{d}}{\text{dx}}(\text{x}^\text{n}\log_\text{a}\text{x})=\text{x}^\text{n}.\frac{1}{\text{x}\log\text{a}}+\log_\text{a}\text{x}(\text{nx}^{\text{n}-1})$
$=\text{x}^{\text{n}-1}\frac{1}{\log\text{a}}+\log_\text{a}\text{x}(\text{nx}^{\text{n}-1})$
$=\text{x}^{\text{n}-1}\Big(\frac{1}{\log\text{a}}+\text{n}\log_{\text{a}}\text{x}\Big)$
View full question & answer→Question 192 Marks
Differentiate the following function with respect to $\text{x}$ $\text{x}^3\sin\text{x}$
AnswerLet $\text{u}=\text{x}^3;\text{v}=\sin\text{x}$Then, $\text{u}'=3\text{x}^2;\text{v}'=\cos\text{x}$
Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$
$\frac{\text{d}}{\text{dx}}(\text{x}^3\sin\text{x})=\text{x}^3\cos\text{x}+\sin\text{x}(3\text{x}^2)$
$=\text{x}^2(\text{x}\cos\text{x}+3\sin\text{x})$
View full question & answer→Question 202 Marks
Find the derivative of f(x) x at x = 1
AnswerWe have,
$\text{f(x)}=\text{x}$
$\because\text{f}'(\text{a})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(a+h)}-\text{f(a)}}{\text{h}}$
$\text{f}'(1)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(1+h)}-\text{f(1)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{(1+h)}-1}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}1$
$\therefore\text{f}'(1)=1$
View full question & answer→Question 212 Marks
Find the derivative of $f(x) = x^2 − 2$ at $x = 10$
AnswerWe have,
$\text{f(x)}=\text{x}^{2}-2$
$\because\text{f}\text{(a)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}\text{(a+h)}-\text{f(a)}}{\text{h}}$
$\text{f}'(10)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(10+h)}-\text{f(10)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{(10+h)}^{2}-2-98}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{100+20\text{h}+\text{h}^{2}-100}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h(20+h)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}(20+\text{h})$
$\therefore\text{f}'(10)=20$
View full question & answer→Question 222 Marks
Differentiate the following function with respect to $\text{x}:$$(\text{x}^3+\text{x}^2+1)\sin\text{x}$
AnswerLet $\text{u}=\text{x}^3+\text{x}^2+1;\text{v}=\sin\text{x}$Then, $\text{u}'=\text{3x}^2+\text{2x};\text{v}'=\cos\text{x}$
Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$
$\frac{\text{d}}{\text{dx}}[(\text{x}^3+\text{x}^2+1)\sin\text{x}]$
$=(\text{x}^3+\text{x}^2+1)\cos\text{x}+(\text{3x}^2+\text{2x})\sin\text{x}$
View full question & answer→Question 232 Marks
2 cosx at x = $\frac{\pi}{2}$
Answer$$We have,
$\because\text{f(x)}=2\cos\text{x}$
$\therefore\text{f}'\text{(a)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{(a+h)}-\text{f(a)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}\Big(\frac{\pi}{2}+\text{h}\Big)-\text{f}\Big(\frac{\pi}{2}\Big)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}\Big(\frac{\pi}{2}+\text{h}\Big)-2\cos\frac{\pi}{2}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-2\text{sinh}-0}{\text{h}}$
$=-2\ \Big[\because\lim_\limits{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$\therefore\text{f}'\Big(\frac{\pi}{2}\Big)=-2$
View full question & answer→Question 242 Marks
Find the derivatives of the following functions at the indicated points:
$\sin\text{x}$ at $\text{x}=\frac{\pi}{2}$
AnswerWe have, $\text{f(x)}=\sin\text{x}$ $\because\text{f}'\text{(a)}\lim\limits_{\text{h}\rightarrow0}\frac{\text{(a+h)}-\text{f(a)}}{\text{h}}$ $\therefore\text{f}'(\frac{\pi}{2})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\frac{\pi}{2}+\text{h})-\text{f}(\frac{\pi}{2})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{sin}(\frac{\pi}{2}+\text{h)}-\sin(\frac{\pi}{2})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{cosh-1}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\Big(1-\frac{\text{h}^{2}}{2!}+\frac{\text{h}^{4}}{4!}\Big)-1}{\text{h}}\ \Big[\because\cos\text{x}=1-\frac{\text{x}^2}{2!}+\frac{\text{x}^4}{4!}-\dots\Big]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\Big(\text{h}\frac{\text{-h}}{2!}+\frac{\text{h}^{3}}{4!}+\frac{\text{h}^{5}}{6!}+...\Big)}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big(-\frac{\text{h}}{2!}+\frac{\text{h}^{3}}{4!}-\frac{\text{h}^{5}}{6!}+...\Big)$ $=0$$\therefore\text{f}'\Big(\frac{\pi}{2}\Big)=0$
View full question & answer→Question 252 Marks
If $\text{y}=\sqrt{\frac{\text{x}}{\text{a}}}+\sqrt{\frac{\text{a}}{\text{x}}},$ prove that $\text{2xy}\frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{x}}{\text{a}}-\frac{\text{a}}{\text{x}}\Big)$
AnswerWe have,$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\sqrt{\frac{\text{x}}{\text{a}}}+\sqrt{\frac{\text{a}}{\text{x}}}\Big)$
$=\frac{1}{\sqrt{\text{a}}}\frac{\text{d}}{\text{dx}}\sqrt{\text{x}}+\sqrt{\text{a}}\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\sqrt{\text{x}}}\Big)$
$=\frac{1}{\sqrt{\text{a}}}\frac{1}{2\sqrt{\text{x}}}+\sqrt{\text{a}}\Big(\frac{-1}{2}\Big)\times\frac{1}{\text{x}\sqrt{\text{x}}}$
$=\frac{1}{\text{2x}}\Bigg(\sqrt{\frac{\text{x}}{\text{a}}}+\Big(-\sqrt{\frac{\text{a}}{\text{x}}}\Big)\Bigg)$
$\Rightarrow\text{2x}\frac{\text{dy}}{\text{dx}}=\sqrt{\frac{\text{x}}{\text{a}}}-\sqrt{\frac{\text{a}}{\text{x}}}$
Mulitiplying both the side by $\text{y}=\sqrt{\frac{\text{x}}{\text{a}}}+\sqrt{\frac{\text{a}}{\text{x}}}$
$\Rightarrow\text{2xy}\frac{\text{dy}}{\text{dx}}=\Big(\sqrt{\frac{\text{x}}{\text{a}}}-\sqrt{\frac{\text{a}}{\text{x}}}\Big)\Big(\sqrt{\frac{\text{x}}{\text{a}}}+\sqrt{\frac{\text{a}}{\text{x}}}\Big)$
$=\Big(\frac{\text{x}}{\text{a}}-\frac{\text{a}}{\text{x}}\Big)$
Hence, proved.
View full question & answer→Question 262 Marks
Differentiate the following function with respect to $(\text{x})$:$3^\text{x}+\text{x}^3+3^3$
AnswerWe have to differentiate $\text{f}(\text{x})$ with respect to $\text{x}$$\frac{\text{d}}{\text{dx}}(3^\text{x}+\text{x}^3+3^3)$
$=\frac{\text{d}}{\text{dx}}(3^\text{x})+\frac{\text{d}}{\text{dx}}(\text{x}^3)+\frac{\text{d}}{\text{dx}}(3^3)$
$=3^\text{x}\log3+3\text{x}^2+0\ \Bigg[\because\frac{\text{d}}{\text{dx}}(\text{a}^\text{x})=\text{a}^\text{x}\log\text{a}\Bigg]$
$=3^\text{x}\log3+3\text{x}^2$
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Differentiate the following functions with respect to x:$\frac{\text{x}^2+1}{\text{x}+1}$
Answer$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2+1}{\text{x}+1}\Big)=\frac{(\text{x}+1)\frac{\text{d}}{\text{dx}}(\text{x}^2+1)\frac{\text{d}}{\text{dx}}(\text{x}+1)}{(\text{x}+1)^2}$$=\frac{(\text{x}+1)\times\text{2x}-(\text{x}^2+1)\times1}{(\text{x}+1)^2}$
$=\frac{\text{2x}^2+\text{2x}-\text{x}^2-1}{(\text{x}+1)^2}$
$=\frac{\text{x}^2+\text{2x}-1}{(\text{x}+1)^2}$
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Differentiate the following function with respect to (x):$\text{x}^4-2\sin\text{x}+\cos\text{x}$
AnswerWe have to differentiate f(x) with respect to x:$\frac{\text{d}}{\text{dx}}(\text{x}^4-2\sin\text{x}+3\cos\text{x})$
$=\frac{\text{d}(\text{x})^4}{\text{dx}}-2\frac{\text{d}}{\text{dx}}(\sin\text{x})+\frac{\text{d}}{\text{dx}}(\cos\text{x})$
$=4\text{x}^3-2\cos\text{x}-3\sin\text{x}$
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Differentiate the following functions with respect to x:$\frac{\text{e}^\text{x}-\tan\text{x}}{\cot\text{x}-\text{x}^\text{n}}$
Answer$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{e}^\text{x}-\tan\text{x}}{\cot\text{x}-\text{x}^\text{n}}\Big)$$=\frac{(\cot\text{x}-\text{x}^\text{n})\frac{\text{d}}{\text{dx}}(\text{e}^\text{x}-\tan\text{x})-(\text{e}^\text{x}-\tan\text{x})\frac{\text{d}}{\text{dx}}(\cot\text{x}-\text{x}^\text{n})}{(\cot\text{x}-\text{x}^\text{n})^2}$
$=\frac{(\cot\text{x}-\text{x}^\text{n})(\text{e}^\text{x}-\sec^2\text{x})-(\text{e}^\text{x}-\tan\text{x})(-\text{cosec}^2\text{x}-\text{nx}^{\text{n}-1})}{(\cot\text{x}-\text{x}^\text{n})^2}$
$=\frac{(\cot\text{x}-\text{x}^\text{n})(\text{e}^\text{x}-\sec^2\text{x})+(\text{e}^\text{x}-\tan\text{x})(\text{cosec}^2\text{x}-\text{nx}^{\text{n}-1})}{(\cot\text{x}-\text{x}^\text{n})^2}$
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Differentiate the following function with respect to $(\text{x})$:$(2\text{x}^2+1)(3\text{x}+2)$
AnswerWe have,$\frac{\text{d}}{\text{dx}}(2\text{x}^2+1)(3\text{x}+2)$
$=(3\text{x}+2)\frac{\text{d}}{\text{dx}}(2\text{x}^2+1)+(\text{2x}^2+1)\frac{\text{d}}{\text{dx}}(3\text{x}+2)$ [Using product rule]
$=(3\text{x}+2)(4\text{x}+0)+(\text{2x}^2+1)(3+0)$
$=(12\text{x}^2+\text{8x}+\text{6x}^2+3)$
$=18\text{x}^2+\text{8x}+3$
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Find the derivative of f (x) = tan x at x = 0
AnswerWe have, $\text{f(x)}=\tan\text{x}$ $\because\text{f}'\text{(a)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(a+h)}-\text{f(a)}}{\text{h}}$ $\therefore\text{f}'(0)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(0+h)}-\text{f(0)}}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\tan \text{h}-\tan0}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{tan h}}{\text{h}}$ $=1\ \Big[\because\lim_\limits{\theta\rightarrow0}\frac{\tan\theta}{\theta}=1\Big]$$\therefore\text{f}'(0)=1$
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Differentiate the following function with respect to x:$\sin^2\text{x}$
Answer$\frac{\text{d}}{\text{dx}}(\sin^2\text{x})$$=2\sin\text{x}\frac{\text{d}}{\text{dx}}(\sin\text{x})$ (Using the chain rule)
$=2\sin\text{x}\cos\text{x}$
$=\sin\text{2x}$
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Differentiate the following functions with respect to x:$\frac{\text{e}^\text{x}+\sin\text{x}}{1+\log\text{x}}$
AnswerWe have, $\frac{\text{d}}{\text{dx}}\Big(\frac{\text{e}^\text{x}+\sin\text{x}}{1+\log\text{x}}\Big)$
Using Quotient rule, we get
$=\frac{(1+\log\text{x})\frac{\text{d}}{\text{dx}}(\text{e}^\text{x}+\sin\text{x})-(\text{e}^\text{x}+\sin\text{x})\frac{\text{d}}{\text{dx}}(1+\log\text{x})}{(1+\log\text{x})^2}$
$=\frac{(1+\log\text{x})(\text{e}^\text{x}+\cos\text{x})-(\text{e}^\text{x}+\sin\text{x})\frac{1}{\text{x}}}{(1+\log\text{x})^2}$
$=\frac{\text{x}(1+\log\text{x})(\text{e}^\text{x}+\cos\text{x})-(\text{e}^\text{x}+\sin\text{x})}{\text{x}(1+\log\text{x})^2}$
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Find the derivative of f(x) = 99x at x = 100
AnswerWe have,
$\text{f}(\text{x}) =99\text{x}$
$\because\text{f}'\text{(a)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(a+h)}-\text{f(a)}}{\text{h}}$
$\text{f}'(100)=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(100+h)-f(100)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{99(100+\text{h})-9900}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{9900+99\text{h}-9900}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0} 99$
$\therefore\text{f}'(100)=99$
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Differentiate the following function with respect to $(\text{x})$:$2\sec\text{x}+3\cot\text{x}-4\tan\text{x}$
AnswerWe have,$\frac{\text{d}}{\text{dx}}(2\sec\text{x}+3\cot\text{x}-4\tan\text{x})$
$=2\frac{\text{d}}{\text{dx}}(\sec\text{x})+3\frac{\text{d}}{\text{dx}}(\cot\text{x})-4\frac{\text{d}}{\text{dx}}(\tan\text{x})$
$=2\sec\text{x}\tan\text{x}-3\text{cosec}^2\text{x}-4\sec^2\text{x}$
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Differentiate the following function with respect to x:
$\sin\text{x}\cos\text{x}$
AnswerLet $\text{u}=\sin\text{x};\text{v}=\cos\text{x}$Then, $\text{u}'=\cos\text{x};\text{v}'=-\sin\text{x}$
Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$
$\frac{\text{d}}{\text{dx}}(\sin\text{x}\cos\text{x})=\sin\text{x}(-\sin\text{x})+\cos\text{x}.\cos\text{x}$
$=-\sin^2\text{x}+\cos^2\text{x}$
$=\cos2\text{x}$
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$\sin2\text{x}$ at $\text{x}=\frac{\pi}{2}$
AnswerWe have, $\text{f}\text{(x)}=\sin2\text{x}$Therefore,
$\text{f}'\text{(a)}=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{a}+\text{h})-\text{f}\text{(a)}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}\Big(\frac{\pi}{2}+\text{h}\Big)-\text{f}\Big(\frac{\pi}{2}\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\Big(\frac{\pi}{2}+\text{h}\Big)-\sin2\Big(\frac{\pi}{2}\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Big(\frac{\pi}{2}\times2+2\text{h}\Big)-\sin(\pi)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{-\cos2\text{h}-0}{\text{h}}$
$=-2$
Therefore, $\text{f}'\Big(\frac{\pi}{2}\Big)=-2$
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Differentiate the following function with respect to $(\text{x})$:$\frac{\text{a}\cos\text{x}+\text{b}\sin\text{x}+\text{c}}{\sin\text{x}}$
AnswerWe have,$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{a}\cos\text{x}+\text{b}\sin\text{x}+\text{c}}{\sin\text{x}}\Big)$
$=\text{a}\frac{\text{d}}{\text{dx}}\Big(\frac{\cos\text{x}}{\sin\text{x}}\Big)+\text{b}\frac{\text{d}}{\text{dx}}(1)+\text{c}\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\sin\text{x}}\Big)$
$=\text{a}(-\text{cosec}^2\text{x})+0+\text{c}(-\text{cosec}\text{x}.\cot\text{x})$
$=-\text{a}\text{cosesc}^2\text{x}-\text{c}\text{cosec}\text{x}.\cot\text{x}$
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Find the derivative of f (x) = cos x at x = 0
Answer$\because\text{f}'(\text{a})=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{f}(\text{a+h})-\text{f}(\text{a})}{\text{h}}$$\text{f}'(0)=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0+\text{h})-\text{f}(0)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\cos(0+\text{h})-\cos0}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\cos\text{h}-\cos0}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\Big(1-\frac{\text{h}^2}{2!}+\frac{\text{h}^4}{4!}\dots\Big)-1}{\text{h}}\ \Big[\because\cos\text{x}=1-\frac{\text{x}^2}{2}+\frac{\text{x}^4}{4!}-\dots\Big]$
$\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}\Big(-\frac{\text{h}}{2!}+\frac{\text{h}^3}{4!}-\frac{\text{h}^5}{6!}-\dots\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\text{h}\Big(-\frac{\text{h}}{2!}+\frac{\text{h}^3}{4!}-\frac{\text{h}^5}{6!}-\dots\Big)$
= 0
$\therefore\text{f}'(0)=0$
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Differentiate the following function with respect to $\text{x}$:$\text{x}^2\text{e}^\text{x}\log\text{x}$
AnswerLet $\text{u}=\text{x}^2;\text{v}=\text{e}^\text{x};\text{w}=\log\text{x}$Then, $\text{u}'=2\text{x};\text{v}'=\text{e}^\text{x};\text{w}=\frac{1}{\text{x}}$
Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uvw})=\text{u}'\text{vw}+\text{u}\text{v}'\text{w}+\text{uv}\text{w}'$
$\frac{\text{d}}{\text{dx}}(\text{x}^2\text{e}^\text{x}\log\text{x})=\text{2xe}^\text{x}\log\text{x}+\text{x}^2\text{e}^\text{x}\log\text{x}+\text{x}^2\text{e}^\text{x}\frac{1}{\text{x}}$
$=\text{2xe}^\text{x}\log\text{x}+\text{x}^2\text{e}^\text{x}\log\text{x}+\text{x}\text{e}^\text{x}$
$=\text{x}\text{e}^\text{x}(2\log\text{x}+\text{x}\log\text{x}+1)$
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