Question

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

Answer 1

Hint: To find the derivative of the given function $y={{\tan }^{2}}\left( 3x \right)$, we will use the chain rule in which we take the derivative by assuming $\tan 3x$ as t and then do the differentiation of ${{\left( t \right)}^{2}}$ in the same way we take the derivative for ${{t}^{n}}$. After that, we will take the derivative of $\tan 3x$ with respect to x which is the same as the differentiation for $\tan x$ with respect to x and then multiply this differentiation with the derivative of $3x$ with respect to x.

Complete step-by-step answer:
The function which we have to take the derivative of is as follows:
$y={{\tan }^{2}}\left( 3x \right)$
We are going to differentiate the above function by assuming $\tan 3x$ as t then we get,
$\Rightarrow y={{t}^{2}}$
Taking differentiation with respect to t on both the sides we get,
$\Rightarrow \dfrac{dy}{dt}=\dfrac{d{{t}^{2}}}{dt}$
Now, differentiating ${{t}^{2}}$ is same as the differentiation of ${{t}^{n}}$ with respect to t which is equal to:
$\dfrac{d\left( {{t}^{n}} \right)}{dt}=n{{t}^{n-1}}$
Substituting n as 2 in the above equation we get,
$\begin{align}
  & \Rightarrow \dfrac{d\left( {{t}^{2}} \right)}{dt}=2{{t}^{2-1}} \\
 & \Rightarrow \dfrac{d\left( {{t}^{2}} \right)}{dt}=2t \\
\end{align}$
Using the above derivative in $\dfrac{dy}{dt}=\dfrac{d{{t}^{2}}}{dt}$ we get,
$\dfrac{dy}{dt}=2t$
Now, substituting $t=\tan 3x$ in the above derivative we get,
$\Rightarrow \dfrac{dy}{dx}=2\left( \tan 3x \right)$
Now, we will take the derivative of $\tan 3x$ with respect to x and then we will multiply this derivative with the above expression on the R.H.S of the equation to get the final derivative of the given function.
Taking the derivative of $\tan 3x$ with respect to x we get,
$\Rightarrow \dfrac{d\tan 3x}{dx}$
The above derivative is achieved by assuming 3x as x and then doing the differentiation. On assuming that we have to find the differentiation for $\tan x$ with respect to x and we know that the differentiation of $\tan x$ with respect to x is equal to:
\[\dfrac{d\tan x}{dx}={{\sec }^{2}}x\]
Now, substituting x as 3x in the above equation we get,
\[\Rightarrow \dfrac{d\tan 3x}{dx}={{\sec }^{2}}3x\]
Now, we are taking the derivative of 3x with respect to x and then multiply this derivative of 3x with respect to x with the above term in the R.H.S we get,
Taking derivative of 3x with respect to x we get,
$\Rightarrow \dfrac{d\left( 3x \right)}{dx}=3$
Multiplying the above derivative with ${{\sec }^{2}}3x$ we get,
$\Rightarrow 3{{\sec }^{2}}3x$
Now, multiplying the above expression by $2\left( \tan 3x \right)$ we get the final derivative of the function given in the above problem is as follows:
$\Rightarrow \dfrac{dy}{dx}=2\left( \tan 3x \right)\left( 3{{\sec }^{2}}3x \right)$
Simplifying the expression written on the R.H.S of the above equation we get,
$\Rightarrow \dfrac{dy}{dx}=6\left( \tan 3x \right)\left( {{\sec }^{2}}3x \right)$
Hence, we have found the derivative of the given function as:
$\dfrac{dy}{dx}=6\left( \tan 3x \right)\left( {{\sec }^{2}}3x \right)$

Note: The mistake that could be possible in the above problem is that while applying the chain rule, you might forget to multiply the derivative of 3x with $2\left( \tan 3x \right)\left( {{\sec }^{2}}3x \right)$ so make sure you won’t make this mistake in the examination and hence, save yourself from the penalization of negative marks in the multiple choice questions.