Question

How do you differentiate the given function ${{\tan }^{2}}\left( 3x \right)$?

Answer 1

Hint: We start solving the problem by assuming \[\tan \left( 3x \right)=z\] 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 $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ to proceed through the problem. We then assuming \[3x=w\] and then make use of 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 facts 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 ${{\tan }^{2}}\left( 3x \right)$.
Let us assume $y={{\tan }^{2}}\left( 3x \right)$ ---(1).
Let us assume \[\tan \left( 3x \right)=z\]. Let us substitute this in equation (1).
$\Rightarrow y={{z}^{2}}$ ---(2).
Let us differentiate both sides of the equation (2) with respect to x.
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( {{z}^{2}} \right)}{dx}$ ---(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).
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( {{z}^{2}} \right)}{dz}\times \dfrac{dz}{dx}$ ---(4).
We know that $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$. Let us use this result in equation (4).
$\Rightarrow \dfrac{dy}{dx}=2z\dfrac{dz}{dx}$ ---(5).
Now, let us substitute \[z=\tan \left( 3x \right)\] in equation (5).
$\Rightarrow \dfrac{dy}{dx}=2\tan \left( 3x \right)\times \dfrac{d\left( \tan \left( 3x \right) \right)}{dx}$ ---(6).
Let us assume \[3x=w\]. Let us substitute this in equation (6).
$\Rightarrow \dfrac{dy}{dx}=2\tan \left( 3x \right)\times \dfrac{d\left( \tan w \right)}{dx}$ ---(7).
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 (7).
$\Rightarrow \dfrac{dy}{dx}=2\tan \left( 3x \right)\times \dfrac{d\left( \tan w \right)}{dw}\times \dfrac{dw}{dx}$ ---(8).
We know that $\dfrac{d\left( \tan x \right)}{dx}={{\sec }^{2}}x$. Let us use this result in equation (8).
$\Rightarrow \dfrac{dy}{dx}=2\tan \left( 3x \right)\times {{\sec }^{2}}w\times \dfrac{dw}{dx}$ ---(9).
Now, let us substitute \[w=3x\] in equation (9).
$\Rightarrow \dfrac{dy}{dx}=2\tan \left( 3x \right)\times {{\sec }^{2}}\left( 3x \right)\times \dfrac{d\left( 3x \right)}{dx}$ ---(10).
We know that $\dfrac{d\left( ax \right)}{dx}=a$. Let us use this result in equation (11).
$\Rightarrow \dfrac{dy}{dx}=2\tan \left( 3x \right)\times {{\sec }^{2}}\left( 3x \right)\times 3$.
$\Rightarrow \dfrac{dy}{dx}=6\tan \left( 3x \right){{\sec }^{2}}\left( 3x \right)$.
$\therefore $ We have found the derivative of the function ${{\tan }^{2}}\left( 3x \right)$ as $6\tan \left( 3x \right){{\sec }^{2}}\left( 3x \right)$.

Note:
We should perform each step carefully in order to avoid confusion and calculation mistakes. Whenever we get this type of problem, we try to make use of chain rules to get a solution to the given problem. We should not forget to different $3x$ after performing equation (5) which is the common mistake done by students. Similarly, we can expect problems to find the derivative of the function $y=\log \left( \sec \left( 5x \right) \right)$.