What is derivative of cotx?

The correct option is –cosec²x (which represents $-{\mathrm{cosec}}^2\left(x\right)$).

To understand why this is correct, we can derive the derivative of $\cot \left(x\right)$ using the quotient rule of differentiation.

First, recall the trigonometric identity for cotangent:
$\cot \left(x\right)=\frac{\cos \left(x\right)}{\sin \left(x\right)}$

Now, let $f\left(x\right)=\frac{u\left(x\right)}{v\left(x\right)}$, where:
$u\left(x\right)=\cos \left(x\right)$ and $v\left(x\right)=\sin \left(x\right)$

Recall the quotient rule for differentiation:
$\frac{d}{dx}\left[\frac{u}{v}\right]=\frac{u\text{'}v-uv\text{'}}{v^2}$

Let's find the derivatives of the individual terms:
$u\text{'}\left(x\right)=\frac{d}{dx}\left[\cos \right(x\left)\right]=-\sin \left(x\right)$
$v\text{'}\left(x\right)=\frac{d}{dx}\left[\sin \right(x\left)\right]=\cos \left(x\right)$

Applying these to the quotient rule:
$\frac{d}{dx}\left[\cot \right(x\left)\right]=\frac{(-\sin (x\left)\right)\left(\sin \right(x\left)\right)-\left(\cos \right(x\left)\right)\left(\cos \right(x\left)\right)}{{\sin }^2\left(x\right)}$

Simplify the numerator:
$\frac{d}{dx}\left[\cot \right(x\left)\right]=\frac{-{\sin }^2\left(x\right)-{\cos }^2\left(x\right)}{{\sin }^2\left(x\right)}$

Factor out the negative sign in the numerator:
$\frac{d}{dx}\left[\cot \right(x\left)\right]=\frac{-\left({\sin }^2\right(x)+{\cos }^2(x\left)\right)}{{\sin }^2\left(x\right)}$

Using the fundamental Pythagorean trigonometric identity, ${\sin }^2\left(x\right)+{\cos }^2\left(x\right)=1$:
$\frac{d}{dx}\left[\cot \right(x\left)\right]=\frac{-1}{{\sin }^2\left(x\right)}$

Since cosecant is the reciprocal of sine ($\mathrm{cosec}\left(x\right)=\frac{1}{\sin \left(x\right)}$):
$\frac{d}{dx}\left[\cot \right(x\left)\right]=-{\mathrm{cosec}}^2\left(x\right)$

Thus, the derivative of $\cot \left(x\right)$ is indeed –cosec²x.