Question

What is the derivative of ${\sec ^2}({x^3})$?

Answer 1

Hint: Here we have to derivate the function ${\sec ^2}({x^3})$. We will derivate the function with the help of chain rule of differentiation which states that the derivative of the composite function can be said as the derivative of the outer function which we multiply by the derivative of the inner function and can be written as $\dfrac{d}{{dx}}f(g(x)) = f'(g(x))g'(x)$.

Complete step by step answer:
Differentiation is used to find the rates of change. It helps us to find the rate of change of the variable $x$ with respect to the variable $y$. The chain rule allows the differentiation of functions which are composite functions and we denote chain rule by $f.g$ where $f$ and $g$ are two functions.

The chain rule of differentiation states that the derivative of the composite function can be said as the derivative of the outer function which we multiply by the derivative of the inner function and can be written as $\dfrac{d}{{dx}}f(g(x)) = f'(g(x))g'(x)$. Here we have to differentiate the function ${\sec ^2}({x^3})$ with respect to $x$.Let us assume,
$y = {\sec ^2}({x^3})$
We can write the above function as $y = {\left( {\sec ({x^3})} \right)^2}$
Now, let $u = {x^3}$ and $t = (\sec u)$, then $y = {t^2}$
differentiating $u = {x^3}$ with respect to $x$. We get,
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{{d({x^3})}}{{dx}}$

We know that $\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$. So,
$ \Rightarrow \dfrac{{d({x^3})}}{{dx}} = 3{x^2}$
Now differentiating $t = (\sec u)$ with respect to $u$. We get,
$ \Rightarrow \dfrac{{dt}}{{du}} = \dfrac{{d(\sec u)}}{{du}}$
We know that $\dfrac{{d(\sec \theta )}}{{d\theta }} = \sec \theta \tan \theta $. So,
$ \Rightarrow \dfrac{{d(\sec u)}}{{du}} = \sec u\tan u$
Now differentiating $y = {t^2}$ with respect to $t$. We get,
$ \Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{{d({t^2})}}{{dt}}$
We know that $\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$. So,
$ \Rightarrow \dfrac{{d({t^2})}}{{dt}} = 2t$

We have to calculate $\dfrac{{dy}}{{dx}}$. So, according to chain rule
$\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dt}} \times \dfrac{{dt}}{{du}} \times \dfrac{{du}}{{dx}}$
Putting $\dfrac{{dy}}{{dt}} = 2t$ ,$\dfrac{{dt}}{{du}} = \sec u\tan u$ and $\dfrac{{du}}{{dx}} = 3{x^2}$. We get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = 2t \times \sec u\tan u \times 3{x^2}$
Substituting $t = (\sec u)$ and $u = {x^3}$. We get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = 2\sec {x^3} \times \sec {x^3}\tan {x^3} \times 3{x^2}$
On multiplying we get,
$ \therefore \dfrac{{dy}}{{dx}} = 6{x^2}{\sec ^2}({x^3})\tan ({x^3})$

Hence, the derivative of the function ${\sec ^2}({x^3})$ is $6{x^2}{\sec ^2}({x^3})\tan ({x^3})$.

Note: Chain rule of differentiation is known as chain rule because we use it to take derivatives of composites of functions and this happens by changing together their derivatives. Note that we use product rule when two functions are being multiplied together and the chain rule is used if the functions are being composed. For example- if we want to find the derivative of the function $f(x) = {x^3}(\cos x)$ we will use product rule, and if we want to the derivative of the function $g(x) = \cos ({x^3})$ we use the chain rule to differentiate it.