Differentiation of cot inverse x

Here you will learn differentiation of cot inverse x or arccotx x by using chain rule.

Let’s begin –

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

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

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

Proof using chain rule :

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

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

\(\implies\) cot y = x

Differentiating both sides with respect to x, we get

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

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

By chain rule,

\(-cosec^2 y\) \(dy\over dx\) = 1

\(dy\over dx\) = \(-1\over cosec^2 y\)

[ \(\because\) 1 + \(cot^2 y\) = \(cosec^2 y\)

\(dy\over dx\) = \(1\over {1 + cot^2 y}\)

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

\(\implies\) \(d\over dx\) \(cot^{-1}x\) = \(-1\over {1 + x^2}\) 

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

Example : What is the differentiation of \(cot^{-1} x^2\) with respect to x ?

Solution : Let y = \(cot^{-1} x^2\)

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

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

\(dy\over dx\) = \(-1\over {1 + x^4}\).(2x) = \(-2x\over {1 + x^4}\)

Hence, \(d\over dx\) (\(cot^{-1} x^2\)) = \(-2x\over {1 + x^4}\)

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

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

Differentiating both sides with respect to x, we get

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

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

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


Related Questions

What is the Differentiation of cot x ?

What is the Integration of Cot Inverse x ?

What is the Differentiation of tan inverse x ?

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