# Differentiation of sec inverse x

Here you will learn differentiation of sec inverse x or arcsecx x by using chain rule.

Let’s begin –

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

If x $$\in$$ R – [-1, 1] . then the differentiation of $$sec^{-1}x$$ with respect to x is $$1\over | x |\sqrt{x^2 – 1}$$.

i.e. $$d\over dx$$ $$sec^{-1}x$$ = $$1\over | x |\sqrt{x^2 – 1}$$.

## Proof using chain rule :

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

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

$$\implies$$ sec y = x

Differentiating both sides with respect to x, we get

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

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

By chain rule,

sec y tan y $$dy\over dx$$ = 1

$$dy\over dx$$ = $$1\over sec y tan y$$

$$dy\over dx$$ = $$1\over | sec y | | tan y |$$

$$dy\over dx$$ = $$1\over | sec y | \sqrt{tan^2 y}$$

$$\implies$$ $$dy\over dx$$ = $$1\over | sec y | \sqrt{sec^2 y – 1}$$

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

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

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

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

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

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

$$dy\over dx$$ = $$1\over | x^2 | \sqrt{x^4 – 1}$$.(2x) = $$2x\over | x^2 | \sqrt{x^4 – 1}$$

Hence, $$d\over dx$$ ($$sec^{-1} x^2$$) = $$2x\over | x^2 | \sqrt{x^4 – 1}$$

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

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

Differentiating both sides with respect to x, we get

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

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

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

### Related Questions

What is the Differentiation of tan inverse x ?

What is the Differentiation of secx ?

What is the Integration of Sec Inverse x and Cosec Inverse x ?