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Last updated date: 09th Dec 2023
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MVSAT Dec 2023

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

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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 $x$is \[2x\]. 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.