How do you find the derivative of the function $y=\tan \left( 5x \right)$?

Hint: We start solving the problem by assuming $5x=z$ and then differentiating both sides of the given function with respect to x. We then recall the chain rule of differentiation as $\dfrac{d\left( g\left( f \right) \right)}{dx}=\dfrac{d\left( g \right)}{df}\times \dfrac{df}{dx}$ to proceed through the problem. We then make use of the fact that $\dfrac{d\left( \tan x \right)}{dx}={{\sec }^{2}}x$ to proceed through the problem. We then make use of the fact that $\dfrac{d\left( ax \right)}{dx}=a$ to get the required answer for the derivative of the function.

Complete step by step answer:
According to the problem, we are asked to find the derivative of the function $y=\tan \left( 5x \right)$.
We have $y=\tan \left( 5x \right)$ -(1).
Let us assume $5x=z$. Let us substitute this in equation (1).
$\Rightarrow y=\tan z$ -(2).
Let us differentiate both sides of the equation (2) with respect to x.
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \tan z \right)}{dx}$ -(3).
From chain rule of differentiation, we know that $\dfrac{d\left( g\left( f \right) \right)}{dx}=\dfrac{d\left( g \right)}{df}\times \dfrac{df}{dx}$. Let us substitute this result in equation (3).
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \tan z \right)}{dz}\times \dfrac{dz}{dx}$ -(4).
We know that $\dfrac{d\left( \tan x \right)}{dx}={{\sec }^{2}}x$. Let us use this result in equation (4).
$\Rightarrow \dfrac{dy}{dx}={{\sec }^{2}}z\times \dfrac{dz}{dx}$ -(5).
Now, let us substitute $z=5x$ in equation (5).
$\Rightarrow \dfrac{dy}{dx}={{\sec }^{2}}\left( 5x \right)\times \dfrac{d\left( 5x \right)}{dx}$ -(6).
We know that $\dfrac{d\left( ax \right)}{dx}=a$. Let us use this result in equation (6).
$\Rightarrow \dfrac{dy}{dx}={{\sec }^{2}}\left( 5x \right)\times 5$.
$\Rightarrow \dfrac{dy}{dx}=5{{\sec }^{2}}\left( 5x \right)$.

$\therefore $ We have found the derivative of the function $y=\tan \left( 5x \right)$ as $5{{\sec }^{2}}\left( 5x \right)$.

Note: Whenever we get this type of problem, we try to make use of chain rule to get a solution to the given problem. We should not forget to different $5x$ after performing equation (5) which is the common mistake done by students. We can also solve this problem by first using the fact $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and then applying $\dfrac{u}{v}$ rule of differentiation $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$ to get the required answer. Similarly, we can expect problems to find the derivative of the function $y=\log \left( \sec \left( 3x \right) \right)$.