Question

How do you take the derivative of $y = {\tan ^2}\left( {{x^3}} \right)$?

Answer 1

In the given problem, we are required to differentiate $y = {\tan ^2}\left( {{x^3}} \right)$ with respect to x. The given function is a composite function, so we will have to apply the chain rule of differentiation in the process of differentiation. So, differentiation of $y = {\tan ^2}\left( {{x^3}} \right)$ with respect to x will be done layer by layer. Also the derivative of $\tan (x)$with respect to x must be remembered.

Complete step by step answer:
So, Derivative of $y = {\tan ^2}\left( {{x^3}} \right)$ with respect to $x$ can be calculated as $\dfrac{d}{{dx}}\left( {{{\tan }^2}\left( {{x^3}} \right)} \right)$. Now,
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left[ {{{\tan }^2}\left( {{x^3}} \right)} \right]$
Now, Let us assume $u = {x^3}$. So substituting $\tan {x^3}$ as $u$, we get,
$\dfrac{d}{{dx}}\left[ {{u^2}} \right]$
Now, we know the product rule of differentiation as,
$\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$.
So, we have,
$2u\left( {\dfrac{{du}}{{dx}}} \right)$
Now, putting back $u$ as $\tan {x^3}$, we get,
$2\tan {x^3}\left( {\dfrac{{d\left( {\tan {x^3}} \right)}}{{dx}}} \right)$

Derivative of tangent function $\tan x$ with respect to x is ${\sec ^2}x$.
So, again following the chain rule of differentiation, we have,
$2\tan {x^3}{\sec ^2}\left( {{x^3}} \right) \times \dfrac{{d\left( {{x^3}} \right)}}{{dx}}$
We know the power rule of differentiation. Hence, we get,
$2\tan {x^3}{\sec ^2}\left( {{x^3}} \right) \times 3{x^2}$
Simplifying the expression,
$6{x^2}\tan {x^3}{\sec ^2}{x^3}$

So, the derivative of $y = {\tan ^2}\left( {{x^3}} \right)$ with respect to $x$ is $6{x^2}\tan {x^3}{\sec ^2}{x^3}$.

Note: The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer. Power rule helps us to differentiate the power functions such as the derivative of ${x^n}$ with respect to x is $n{x^{n - 1}}$.