Question

What is the derivative of \[y = \sec \left( {{x^2}} \right)\] ?

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

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.