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 ?

Leave a Comment

Your email address will not be published.