Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

$\log (\operatorname{cosec} x-\cot x)$

Solution:

Let $y=\log (\operatorname{cosec} x-\cot x)$

On differentiating $y$ with respect to $x$, we get

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

We know $\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})=\frac{1}{\mathrm{x}}$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\operatorname{cosec} x-\cot x} \frac{d}{d x}(\operatorname{cosec} x-\cot x)$ [using chain rule]

$\Rightarrow \frac{d y}{d x}=\frac{1}{\operatorname{cosec} x-\cot x}\left[\frac{d}{d x}(\operatorname{cosec} x)-\frac{d}{d x}(\cot x)\right]$

We know $\frac{d}{d x}(\operatorname{cosec} x)=-\operatorname{cosec} x \cot x$ and $\frac{d}{d x}(\cot x)=-\operatorname{cosec}^{2} x$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\operatorname{cosec} x-\cot x}\left[-\operatorname{cosec} x \cot x-\left(-\operatorname{cosec}^{2} x\right)\right]$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\operatorname{cosec} x-\cot x}\left[-\operatorname{cosec} x \cot x+\operatorname{cosec}^{2} x\right]$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\operatorname{cosec} x-\cot x}\left[\operatorname{cosec}^{2} x-\operatorname{cosec} x \cot x\right]$

$\Rightarrow \frac{d y}{d x}=\frac{1}{\operatorname{cosec} x-\cot x}[(\operatorname{cosec} x-\cot x) \operatorname{cosec} x]$

$\therefore \frac{d y}{d x}=\operatorname{cosec} x$

Thus, $\frac{d}{d x}[\log (\operatorname{cosec} x-\cot x)]=\operatorname{cosecx}$