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Question

Question: What is the derivative of \[\tan {x^2}\] ?...

What is the derivative of tanx2\tan {x^2} ?

Explanation

Solution

Hint : Here we need to differentiate the given problem with respect to x. We know that the differentiation of xn{x^n} with respect to ‘x’ is d(xn)dx=n.xn1\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}} . We take u=x2u = {x^2} and then we differentiate it with respect to x.

Complete step by step solution:
Given,
tanx2\tan {x^2} .
Let put u=x2u = {x^2}, then
tanx2=tanu\tan {x^2} = \tan u
Now differentiating with respect to ‘x’ we have,
ddx(tanx2)=ddx(tanu)\dfrac{d}{{dx}}\left( {\tan {x^2}} \right) = \dfrac{d}{{dx}}\left( {\tan u} \right)
We know the differentiation of tangent function,
ddx(tanx2)=sec2(u)ddx(u)\dfrac{d}{{dx}}\left( {\tan {x^2}} \right) = {\sec ^2}\left( u \right)\dfrac{d}{{dx}}\left( u \right)
But we have u=x2u = {x^2} then,
ddx(tanx2)=sec2(x2)ddx(x2)\dfrac{d}{{dx}}\left( {\tan {x^2}} \right) = {\sec ^2}\left( {{x^2}} \right)\dfrac{d}{{dx}}\left( {{x^2}} \right)
ddx(tanx2)=sec2(x2).2x\dfrac{d}{{dx}}\left( {\tan {x^2}} \right) = {\sec ^2}\left( {{x^2}} \right).2x
Thus we have,
ddx(tanx2)=2x.sec2(x2)\Rightarrow \dfrac{d}{{dx}}\left( {\tan {x^2}} \right) = 2x.{\sec ^2}\left( {{x^2}} \right) . This is the required result.
So, the correct answer is “ 2x.sec2(x2)2x.{\sec ^2}\left( {{x^2}} \right) ”.

Note : We know the differentiation of xn{x^n} is d(xn)dx=n.xn1\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}} . The obtained result is the first derivative. If we differentiate again we get a second derivative. If we differentiate the second derivative again we get a third derivative and so on. Careful in applying the product rule. We also know that differentiation of constant terms is zero.