# Differentiation of cot inverse x

Here you will learn differentiation of cot inverse x or arccotx x by using chain rule.

Let’s begin –

## Differentiation of cot inverse x or $$cot^{-1}x$$ :

The differentiation of $$cot^{-1}x$$ with respect to x is $$-1\over {1 + x^2}$$.

i.e. $$d\over dx$$ $$cot^{-1}x$$ = $$-1\over {1 + x^2}$$.

## Proof using chain rule :

Let y = $$cot^{-1}x$$. Then,

$$cot(cot^{-1}x)$$ = x

$$\implies$$ cot y = x

Differentiating both sides with respect to x, we get

$$d\over dx$$(cot y) = $$d\over dx$$(x)

$$d\over dx$$ (cot y) = 1

By chain rule,

$$-cosec^2 y$$ $$dy\over dx$$ = 1

$$dy\over dx$$ = $$-1\over cosec^2 y$$

[ $$\because$$ 1 + $$cot^2 y$$ = $$cosec^2 y$$

$$dy\over dx$$ = $$1\over {1 + cot^2 y}$$

$$\implies$$ $$dy\over dx$$ = $$-1\over {1 + x^2}$$

$$\implies$$ $$d\over dx$$ $$cot^{-1}x$$ = $$-1\over {1 + x^2}$$

Hence, the differentiation of $$cot^{-1}x$$ with respect to x is $$-1\over {1 + x^2}$$.

Example : What is the differentiation of $$cot^{-1} x^2$$ with respect to x ?

Solution : Let y = $$cot^{-1} x^2$$

Differentiating both sides with respect to x and using chain rule, we get

$$dy\over dx$$ = $$d\over dx$$ ($$cot^{-1} x^2$$)

$$dy\over dx$$ = $$-1\over {1 + x^4}$$.(2x) = $$-2x\over {1 + x^4}$$

Hence, $$d\over dx$$ ($$cot^{-1} x^2$$) = $$-2x\over {1 + x^4}$$

Example : What is the differentiation of x + $$cot^{-1} x$$ with respect to x ?

Solution : Let y = x + $$cot^{-1} x$$

Differentiating both sides with respect to x, we get

$$dy\over dx$$ = $$d\over dx$$ (x) + $$d\over dx$$ ($$cot^{-1} x$$)

$$dy\over dx$$ = 1 + $$-1\over {1 + x^2}$$

Hence, $$d\over dx$$ (x + $$cot^{-1} x$$) = 1 – $$1\over {1 + x^2}$$

### Related Questions

What is the Differentiation of cot x ?

What is the Integration of Cot Inverse x ?

What is the Differentiation of tan inverse x ?