# Differentiation of sin inverse x

Here you will learn differentiation of sin inverse x or arcsin x by using chain rule.

Let’s begin –

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

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

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

## Proof using chain rule :

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

$$sin(sin^{-1}x)$$ = x $$\implies$$ sin y = x

Differentiating both sides with respect to x, we get

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

By chain rule,

1 = $$d\over dx$$ (sin y) $$\times$$ $$dy\over dx$$

1 = cos y $$dy\over dx$$

$$dy\over dx$$ = $$1\over cos y$$

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

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

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

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

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

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

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

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

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

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

Example : What is the differentiation of $$x^2$$ + $$sin^{-1} x^5$$ with respect to x ?

Solution : Let y = $$x^2$$ + $$sin^{-1} x^5$$

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

$$dy\over dx$$ = $$d\over dx$$ ($$x^2$$) + $$d\over dx$$ ($$sin^{-1} x^5$$)

$$dy\over dx$$ = 2x + $$1\over \sqrt{1 – x^{10}}$$.$$5x^4$$

Hence, $$d\over dx$$ ($$x^2$$ + $$sin^{-1} x^5$$)= 2x + $$5x^4\over \sqrt{1 – x^{10}}$$

### Related Questions

What is the Differentiation of sin x ?

What is the Integration of Sin Inverse x ?

What is the Differentiation of sec inverse x ?