# Differentiation of cos inverse x

Here you will learn differentiation of cos inverse x or arccos x by using chain rule.

Let’s begin –

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

If x $$\in$$ (-1, 1) , then the differentiation of $$cos^{-1}x$$ with respect to x is $$-1\over \sqrt{1 – x^2}$$.

i.e. $$d\over dx$$ $$cos^{-1}x$$ = $$-1\over \sqrt{1 – x^2}$$ , for x $$\in$$ (-1, 1).

## Proof using chain rule :

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

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

$$\implies$$ cos y = x

Differentiating both sides with respect to x, we get

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

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

By chain rule,

-sin y $$dy\over dx$$ = 1

$$dy\over dx$$ = $$-1\over sin y$$

$$dy\over dx$$ = $$-1\over \sqrt{1 – cos^2 y}$$

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

$$\implies$$ $$d\over dx$$ $$cos^{-1}x$$ = $$-1\over \sqrt{1 – x^2}$$

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

Example : What is the differentiation of $$cos^{-1} x^3$$ with respect to x ?

Solution : Let y = $$cos^{-1} x^3$$

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

$$dy\over dx$$ = $$d\over dx$$ ($$cos^{-1} x^3$$)

$$dy\over dx$$ = $$-1\over \sqrt{1 – x^6}$$.$$3x^2$$ = $$-3x^2\over \sqrt{1 – x^6}$$

Hence, $$d\over dx$$ ($$cos^{-1} x^3$$) = $$-3x^2\over \sqrt{1 – x^6}$$

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

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

Differentiating both sides with respect to x, we get

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

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

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

### Related Questions

What is the Differentiation of cos x ?

What is the Integration of cos Inverse x ?

What is the Differentiation of cosec inverse x ?