# Differentiation of Inverse Trigonometric Functions

Here you will learn what is the differentiation of inverse trigonometric functions with examples.

Let’s begin –

## Differentiation of Inverse Trigonometric Functions

(i)  If x $$\in$$ (-1, 1), then the differentiation of $$sin^{-1}x$$ or arcsinx 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).

(ii)  If x $$\in$$ (-1, 1), then the differentiation of $$cos^{-1}x$$ or arccosx 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).

(iii)  The differentiation of $$tan^{-1}x$$ or arctanx with respect to x is $$1\over {1+x^2}$$.

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

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

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

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

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

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

i.e. $$d\over dx$$ $$cosec^{-1}x$$ = $$-1\over |x|\sqrt{x^2-1}$$ , x $$\in$$ R – [-1, 1]

Example 1 : find the differentiation of $$sin^{-1}5x$$.

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

Now, $$dy\over dx$$ = $$1\over \sqrt{1-(5x)^2}$$.5

= $$5\over \sqrt{1-25x^2}$$

Example 2 : find the differentiation of $$cos^{-1}5x$$.

Solution : Let y = $$cos^{-1}5x$$

Now, $$dy\over dx$$ = $$-1\over \sqrt{1-(5x)^2}$$.5

= $$-5\over \sqrt{1-25x^2}$$

Example 3 : find the differentiation of $$tan^{-1}5x$$.

Solution : Let y = $$tan^{-1}5x$$

Now, $$dy\over dx$$ = $$1\over {1+(5x)^2}$$.5

= $$5\over {1+25x^2}$$

Example 4 : find the differentiation of $$sec^{-1}5x$$.

Solution : Let y = $$sec^{-1}5x$$

Now, $$dy\over dx$$ = $$1\over |5x|\sqrt{(5x)^2 – 1}$$.5

= $$5\over |5x|\sqrt{25x^2-1}$$ = $$1\over x\sqrt{25x^2-1}$$

Example 5 : find the differentiation of $$cosec^{-1}5x$$.

Solution : Let y = $$cosec^{-1}5x$$

Now, $$dy\over dx$$ = $$-1\over |5x|\sqrt{1-(5x)^2}$$.5

= $$-5\over |5x|\sqrt{25x^2-1}$$ = $$-1\over x\sqrt{25x^2-1}$$