Question
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 $\text{f}'(\text{x})=-3\text{cosec}^2-5\text{cosecx}\cot\text{x}$
$=\Bigg(-\lim\limits_{{\text{h}}\rightarrow0}\frac{\sin\text{h}}{\text{h}}\Bigg).\Bigg(\lim\limits_{{\text{h}}\rightarrow0}(\frac{1}{\sin\text{x}.\sin(\text{x}+\text{h})}\Bigg)$
$=-1.\frac{1}{\sin\text{x}.\sin(\text{x}+0)}=\frac{-1}{\sin^2\text{x}}=-\text{cosec}^2\text{x}$ ...(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 $\text{f}'(\text{x})=-3\text{cosec}^2-5\text{cosecx}\cot\text{x}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find the linear inequations for which the shaded area in Fig. is the solution set. Draw the diagram of the solution set of the linear inequations:
Find the equations to the altitudes of the triangle whose angular points are A(2, -2), B(1, 1) and C(-1, 0).
If $z_1, z_2$ are two complex numbers such that $|\text{z}_1|=|\text{z}_2|$ and $\text{arg(z}_1)+\text{arg(z}_2)=\pi,$ then show that $\text{z}_1=-\bar{\text{z}}_2.$
→Determine the number of 5 cards combinations out of a deck of 52 cards if at least one of the 5 cards has to be a king?
Prove that: $\sin10^\circ\sin50^\circ\sin60^\circ\sin70^\circ=\frac{\sqrt3}{16}$
→Solve the following equation: $\sqrt{3}\text{x}^2-\sqrt{2}\text{x}+3\sqrt{3}=0.$
→Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}+\tan^2\text{x}}{\text{x}\sin\text{x}}$
→Find the equations of two straight lines passing through $(1, 2)$ and making an angle of $60°$ with the line $x + y = 0$. Find also the area of the triangle formed by the three lines.
→Differentiate the following functions with respect to x:$\frac{\sin\text{x}-\text{x}\cos\text{x}}{\text{x}\sin\text{x}+\cos\text{x}}$
→Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.