(Each question 4 marks)

Question 514 Marks

Find the derivative of cos x from first principle.

Answer

Let $ \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]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{\cos\text{x}-\cos(\text{x}+\text{h})}{\cos\text{x}\cos(\text{x}+\text{h})}\bigg]$ $=\frac{1}{\cos\text{x}}.\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{-2\sin\big(\frac{\text{x}+\text{x}+\text{h}}{2}\big)\sin\big(\frac{\text{x}-\text{x}-\text{h}}{2}\big)}{\cos(\text{x}+\text{h})}\bigg]$ $=\frac{1}{\cos\text{x}}.\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{-2\sin\big(\frac{2\text{x}+\text{h}}{2}\big)\sin\big(-\frac{\text{h}}{2}\big)}{\cos(\text{x}+\text{h})}\bigg]$ $=\frac{1}{\cos\text{x}}.\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\frac{\Bigg[-\sin\big(\frac{2\text{x}+\text{h}}{2}\big)\frac{\sin\big(\frac{\text{h}}{2}\big)}{\big(\frac{\text{h}}{2}\big)}\Bigg]}{\cos(\text{x}+\text{h})}$ $=\frac{1}{\cos\text{x}}.\lim\limits_{\frac{\text{h}}{{2}}\rightarrow0}\frac{\sin\bigg(\frac{\text{h}}{2}\bigg)}{\bigg(\frac{\text{h}}{2}\bigg)}.\lim\limits_{\text{h}\rightarrow0}\frac{\sin\bigg(\frac{2\text{x}+\text{h}}{2}\bigg)}{\cos(\text{x}+\text{h})}$ $=\frac{1}{\cos\text{x}}.1.\frac{\sin\text{x}}{\cos\text{x}}$ $=\sec\text{x}\tan\text{x}$

View full question & answer→

Question 534 Marks

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)$.

Answer

The 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,$\text{f}'(1)=100\times\text{f}'(0)$

View full question & answer→

Question 544 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): $(\text{x}+\cos\text{x})(\text{x}-\tan\text{x})$

Answer

Let $\text{f(x)}=(\text{x}+\cos\text{x})(\text{x}-\tan\text{x})$ 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:$\text{f}(\text{x}) =5\sec\text{x}+4\cos\text{x}$

Answer

Let $\text{f}(\text{x}) =5\sec\text{x}+4\cos\text{x}$. 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): $\frac{\sin{(\text{x+a)}}}{\cos\text{x}}$

Answer

Let $\text{f(x)}=\frac{\sin{(\text{x+a})}}{\cos\text{x}}$ 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: $\text{cosecx}$

Answer

View full question & answer→

Question 584 Marks

Find the derivative of the following functions: $ 2 \tan\text{x} – 7\sec\text{x}$

Answer

Let $\text{f}(\text{x})= 2 \tan\text{x} – 7\sec\text{x}$ 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: $5\sin\text{x} – 6\cos\text{x} + 7$

Answer

Let $\text{f}(\text{x})=5\sin\text{x} – 6\cos\text{x} + 7$ 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: $3\cot\text{x} + 5\text{cosec}\text{x}$

Answer

Let $\text{f}(\text{x})=3\cot\text{x} + 5\text{cosec}\text{x}$. 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 $\text{f}'(\text{x})=-3\text{cosec}^2-5\text{cosecx}\cot\text{x}$

View full question & answer→

MATHS (Each question 4 marks)