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

How do you differentiate arctan(x2)\arctan \left( {{x^2}} \right)?

Explanation

Solution

In the given problem, we are required to differentiate arctan(x2)\arctan \left( {{x^2}} \right) with respect to x. Since, arctan(x2)\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(x2)\arctan \left( {{x^2}} \right) . So, differentiation of arctan(x2)\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(x2)\arctan \left( {{x^2}} \right) with respect to xx can be calculated as ddx(arctan(x2))\dfrac{d}{{dx}}\left( {\arctan \left( {{x^2}} \right)} \right) .
We know that arctan(x)\arctan \left( x \right) is the same as tan1(x)ta{n^{ - 1}}\left( x \right). So, we get,
ddx(tan1(x2))\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 ddx(tan1(x2))\dfrac{d}{{dx}}\left( {ta{n^{ - 1}}\left( {{x^2}} \right)} \right)
Now, Let us assume u=x2u = {x^2}. So substituting x2{x^2} as uu, we get,
\Rightarrow $$$\dfrac{d}{{dx}}\left[ {{{\tan }^{ - 1}}u} \right]$$ Now, we know that the derivative ofta{n^{ - 1}}\left( x \right)withrespecttoxiswith respect to x is\dfrac{1}{{1 + {x^2}}}.So,weget,. So, we get, \Rightarrow \dfrac{1}{{1 + {u^2}}}\left( {\dfrac{{du}}{{dx}}} \right)Now,puttingback Now, putting backuasas {x^2},weget,, 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}withrespecttowith respect toxis $$2x$$. So, \dfrac{d}{{dx}}\left( {{x^2}} \right) = 2x.So,Substitutingtheequivalentexpressionof. So, Substituting the equivalent expression of \dfrac{d}{{dx}}\left( {{x^2}} \right),weget,, we get, \Rightarrow \dfrac{1}{{1 + {x^4}}} \times 2xSimplifyingtheexpression,weget, Simplifying the expression, we get, \Rightarrow \dfrac{{2x}}{{1 + {x^4}}}So,thederivativeof So, the derivative of\arctan \left( {{x^2}} \right)withrespecttowith respect toxisis\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.