Differentiation of cos inverse x

Here you will learn differentiation of cos inverse x or arccos x by using chain rule.

Let’s begin –

Differentiation of cos inverse x or \(cos^{-1}x\) :

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

Proof using chain rule :

Let y = \(cos^{-1}x\). Then,

\(cos(cos^{-1}x)\) = x

\(\implies\) cos y = x

Differentiating both sides with respect to x, we get

\(d\over dx\)(cos y) = \(d\over dx\)(x)

\(d\over dx\) (cos y) = 1

By chain rule,

-sin y \(dy\over dx\) = 1

\(dy\over dx\) = \(-1\over sin y\)

\(dy\over dx\) = \(-1\over \sqrt{1 – cos^2 y}\)

\(\implies\) \(dy\over dx\) = \(-1\over \sqrt{1 – x^2}\)

\(\implies\) \(d\over dx\) \(cos^{-1}x\) = \(-1\over \sqrt{1 – x^2}\) 

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

Example : What is the differentiation of \(cos^{-1} x^3\) with respect to x ?

Solution : Let y = \(cos^{-1} x^3\)

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

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

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

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

Example : What is the differentiation of x + \(cos^{-1} x\) with respect to x ?

Solution : Let y = x + \(cos^{-1} x\)

Differentiating both sides with respect to x, we get

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

\(dy\over dx\) = 1 + \(-1\over \sqrt{1 – x^2}\)

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


Related Questions

What is the Differentiation of cos x ?

What is the Integration of cos Inverse x ?

What is the Differentiation of cosec inverse x ?

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