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.