Question

If $y=\ln (cosecx-\cot x)$ then $\frac{dy}{dx}$ equals to

• A. $cosecx+\cot x$
• B. $\cot x$
• C. $\sec x+\tan x$
• D. $cosecx$

Answer 1

Answer: (D)

$cosecx$

$y=\ln (cosecx-\cot x)$

Chain rule,

$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$

Here $u=cosecx-\cot x$

$\Rightarrow \frac{dy}{dx}$$=\frac{1}{cosecx-\cot x}\cdot \frac{d(cosecx-\cot x)}{dx}$

$\Rightarrow \frac{dy}{dx}$$=\frac{-cosecx.\cot x+cosec{}^{2}x}{cosecx-\cot x}$

$\Rightarrow \frac{dy}{dx}$$=\frac{-cosecx(\cot x-cosecx)}{cosecx-\cot x}$

$\Rightarrow \frac{dy}{dx}$$=cosecx$