Differentiation of tan inverse x

Here you will learn differentiation of tan inverse x or arctanx x by using chain rule.

Let’s begin –

Differentiation of tan inverse x or $tan^{-1}x$ :

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

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

Proof using chain rule :

Let y = $tan^{-1}x$. Then,

$tan(tan^{-1}x)$ = x

$\implies$ tan y = x

Differentiating both sides with respect to x, we get

$d\over dx$(tan y) = $d\over dx$(x)

$d\over dx$ (tan y) = 1

By chain rule,

$sec^2 y$ $dy\over dx$ = 1

$dy\over dx$ = $1\over sec^2 y$

[ $\because$ 1 + $tan^2 y$ = $sec^2 y$

$dy\over dx$ = $1\over {1 + tan^2 y}$

$\implies$ $dy\over dx$ = $1\over {1 + x^2}$

$\implies$ $d\over dx$ $tan^{-1}x$ = $1\over {1 + x^2}$ 

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

Example : What is the differentiation of $tan^{-1} x^2$ with respect to x ?

Solution : Let y = $tan^{-1} x^2$

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

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

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

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

Example : What is the differentiation of 2x + $tan^{-1} x$ with respect to x ?

Solution : Let y = 2x + $tan^{-1} x$

Differentiating both sides with respect to x, we get

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

$dy\over dx$ = 2 + $1\over {1 + x^2}$

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

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Related Questions

What is the Differentiation of tanx ?

What is the Integration of Tan Inverse x ?

What is the Differentiation of cos inverse x ?