Question

Differentiate w.r.t x: log(cosecx-cotx)

Answer 1

Please check the attachment.

Answer 2

Answer:

$\frac{\mathrm{dy}}{\mathrm{dx}}(\log (\csc x-\cot x))=\csc x$

Step-by-step explanation:

Let y = log (cosec x- cot x)

We know that, the general formula for $\log x=\frac{1}{x}$

Now applying the above formula and differentiating both sides with respect to x, we get

$\frac{d y}{d x}=\frac{d y}{d x}[\log (\csc x-\cot x)]$

Now, by applying chain differentiation to the above step, we get  

$=\frac{1}{\csc x-\cot x}\left[-\csc x . \cot x-\left(-\csc ^{2} x\right)\right]$

On simplifying the above step, we get,

$=\frac{-\csc x \cot x+\csc ^{2} x}{\csc x-\cot x}$

From the above step,  csc x is taken as common in the numerator, we get

$=\frac{\csc x(\csc x-\cot x)}{\csc x-\cot x}$

$\therefore \frac{d y}{d x}=\csc x$