Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# How do you differentiate $\arctan \left( {{x^2}} \right)$?

Last updated date: 02nd Aug 2024
Total views: 352.8k
Views today: 8.52k
Verified
352.8k+ views
Hint: In the given problem, we are required to differentiate $\arctan \left( {{x^2}} \right)$ with respect to x. Since, $\arctan \left( {{x^2}} \right)$ is a composite function, so we will have to apply chain rule of differentiation in the process of differentiating $\arctan \left( {{x^2}} \right)$ . So, differentiation of $\arctan \left( {{x^2}} \right)$ with respect to x will be done layer by layer using the chain rule of differentiation.

Complete step-by-step solution:
So, Derivative of $\arctan \left( {{x^2}} \right)$ with respect to $x$ can be calculated as $\dfrac{d}{{dx}}\left( {\arctan \left( {{x^2}} \right)} \right)$ .
We know that $\arctan \left( x \right)$ is the same as $ta{n^{ - 1}}\left( x \right)$. So, we get,
$\dfrac{d}{{dx}}\left( {ta{n^{ - 1}}\left( {{x^2}} \right)} \right)$
Taking the power outside the bracket in order to apply chain rule of differentiation.
$\Rightarrow $$\dfrac{d}{{dx}}\left( {ta{n^{ - 1}}\left( {{x^2}} \right)} \right) Now, Let us assume u = {x^2}. So substituting {x^2} as u, we get, \Rightarrow $\dfrac{d}{{dx}}\left[ {{{\tan }^{ - 1}}u} \right]$ Now, we know that the derivative of ta{n^{ - 1}}\left( x \right) with respect to x is \dfrac{1}{{1 + {x^2}}}. So, we get, \Rightarrow$$\dfrac{1}{{1 + {u^2}}}\left( {\dfrac{{du}}{{dx}}} \right)$
Now, putting back $u$as ${x^2}$, we get,
$\Rightarrow $$\dfrac{1}{{1 + {{\left( {{x^2}} \right)}^2}}}\left( {\dfrac{{d\left( {{x^2}} \right)}}{{dx}}} \right) because $\dfrac{{du}}{{dx}} = \dfrac{{d({x^2})}}{{dx}}$ Now, we know that derivative of {x^2} with respect to xis $2x$. So, \dfrac{d}{{dx}}\left( {{x^2}} \right) = 2x. So, Substituting the equivalent expression of \dfrac{d}{{dx}}\left( {{x^2}} \right), we get, \Rightarrow$$\dfrac{1}{{1 + {x^4}}} \times 2x$
Simplifying the expression, we get,
$\Rightarrow$$\dfrac{{2x}}{{1 + {x^4}}}$
So, the derivative of $\arctan \left( {{x^2}} \right)$ with respect to $x$ is $\dfrac{{2x}}{{1 + {x^4}}}$.

Note: The given problem may also be solved using the first principle of differentiation. 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.