Question

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

Answer 1

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

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

Note: We know the differentiation of \[{x^n}\] is \[\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}\] . 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.