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

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

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