Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

What is the derivative of ${\tan ^{ - 1}}\left( {{x^2}} \right)$?

Answer
VerifiedVerified
455.4k+ views
Hint: In the given problem, we are required to differentiate ${\tan ^{ - 1}}\left( {{x^2}} \right)$ with respect to x. The given function is a composite function, so we will have to apply the chain rule of differentiation in the process of differentiation. So, differentiation of ${\tan ^{ - 1}}\left( {{x^2}} \right)$ with respect to x will be done layer by layer. Also the derivative of inverse tangent function must be remembered.

Complete answer:
So, Derivative of ${\tan ^{ - 1}}\left( {{x^2}} \right)$ with respect to x can be calculated as $\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( {{x^2}} \right)} \right)$ .
Now, $\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( {{x^2}} \right)} \right)$
Now, Let us assume $u = {x^2}$. So substituting ${x^2}$ as $u$, we get,
$ = $$\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}u} \right)$
Now, we know that the derivative of the inverse tangent function ${\tan ^{ - 1}}y$ with respect to y is $\left( {\dfrac{1}{{1 + {y^2}}}} \right)$.
So, we get,
$ = \left( {\dfrac{1}{{1 + {u^2}}}} \right)\left( {\dfrac{{du}}{{dx}}} \right)$
Now, we substitute back the value of u in terms of x as $u = {x^2}$.
 So, we get,
$ = \left( {\dfrac{1}{{1 + {{\left( {{x^2}} \right)}^2}}}} \right)\left( {\dfrac{{d\left( {{x^2}} \right)}}{{dx}}} \right)$
Simplifying the expression, we get,
$ = \left( {\dfrac{1}{{1 + {x^4}}}} \right) \times \dfrac{{d\left( {{x^2}} \right)}}{{dx}}$
Now, we know the power rule of differentiation as $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$.
So, we know that the derivative of ${x^n}$ with respect to $x$ is \[n{x^{n - 1}}\]. Hence, we get,
$ = \left( {\dfrac{1}{{1 + {x^4}}}} \right) \times \left( {2{x^{2 - 1}}} \right)$
Simplifying further, we get,
$ = \left( {\dfrac{{2x}}{{1 + {x^4}}}} \right)$
So, the derivative of ${\tan ^{ - 1}}\left( {{x^2}} \right)$ with respect to $x$ is $\left( {\dfrac{{2x}}{{1 + {x^4}}}} \right)$.

Note:
The derivatives of basic trigonometric functions must be learned by heart in order to find derivatives of complex composite functions using chain rule of differentiation. The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer. Remember to substitute back the assigned variable in terms of the original variable given in the question.