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Question

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

What is the derivative of tan1(x2){\tan ^{ - 1}}\left( {{x^2}} \right)?

Explanation

Solution

In the given problem, we are required to differentiate tan1(x2){\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 tan1(x2){\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 tan1(x2){\tan ^{ - 1}}\left( {{x^2}} \right) with respect to x can be calculated as ddx(tan1(x2))\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( {{x^2}} \right)} \right) .
Now, ddx(tan1(x2))\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( {{x^2}} \right)} \right)
Now, Let us assume u=x2u = {x^2}. So substituting x2{x^2} as uu, we get,
== ddx(tan1u)\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}u} \right)
Now, we know that the derivative of the inverse tangent function tan1y{\tan ^{ - 1}}y with respect to y is (11+y2)\left( {\dfrac{1}{{1 + {y^2}}}} \right).
So, we get,
=(11+u2)(dudx)= \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=x2u = {x^2}.
So, we get,
=(11+(x2)2)(d(x2)dx)= \left( {\dfrac{1}{{1 + {{\left( {{x^2}} \right)}^2}}}} \right)\left( {\dfrac{{d\left( {{x^2}} \right)}}{{dx}}} \right)
Simplifying the expression, we get,
=(11+x4)×d(x2)dx= \left( {\dfrac{1}{{1 + {x^4}}}} \right) \times \dfrac{{d\left( {{x^2}} \right)}}{{dx}}
Now, we know the power rule of differentiation as d(xn)dx=nxn1\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}.
So, we know that the derivative of xn{x^n} with respect to xx is nxn1n{x^{n - 1}}. Hence, we get,
=(11+x4)×(2x21)= \left( {\dfrac{1}{{1 + {x^4}}}} \right) \times \left( {2{x^{2 - 1}}} \right)
Simplifying further, we get,
=(2x1+x4)= \left( {\dfrac{{2x}}{{1 + {x^4}}}} \right)
So, the derivative of tan1(x2){\tan ^{ - 1}}\left( {{x^2}} \right) with respect to xx is (2x1+x4)\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.