Question

Find the derivative of ${{x}^{2}}{{e}^{3x}}{{\tan }^{4}}x$ .

Answer 1

Hint: To find the derivative of the given function, we have to apply the product rule for three functions which is given by the formula $\dfrac{d}{dx}\left( uvw \right)={u}'vw+u{v}'w+uv{w}'$ . To find the derivative of ${{x}^{2}}$ , ${{e}^{3x}}$ and ${{\tan }^{4}}x$ after substituting in the product rule formula, we have to use standard derivative results and chain rule.

Complete step by step answer:
We have to differentiate ${{x}^{2}}{{e}^{3x}}{{\tan }^{4}}x$ . We can write this mathematically as
$\Rightarrow \dfrac{d}{dx}\left( {{x}^{2}}{{e}^{3x}}{{\tan }^{4}}x \right)$
We have to apply product rules. We know that $\dfrac{d}{dx}\left( uvw \right)={u}'vw+u{v}'w+uv{w}'$ . Therefore, we can write the derivative of the given function as
\[\Rightarrow \dfrac{d}{dx}\left( {{x}^{2}}{{e}^{3x}}{{\tan }^{4}}x \right)={{e}^{3x}}{{\tan }^{4}}x\times \dfrac{d}{dx}\left( {{x}^{2}} \right)+{{x}^{2}}{{\tan }^{4}}x\times \dfrac{d}{dx}\left( {{e}^{3x}} \right)+{{x}^{2}}{{e}^{3x}}\dfrac{d}{dx}\left( {{\tan }^{4}}x \right)...\left( i \right)\]
Let us find the derivatives of the above functions. Let us consider \[{{e}^{3x}}{{\tan }^{4}}x\times \dfrac{d}{dx}\left( {{x}^{2}} \right)\] .
We know that $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ . Therefore, we can write the above function as
 \[\begin{align}
  & \Rightarrow {{e}^{3x}}{{\tan }^{4}}x\times \dfrac{d}{dx}\left( {{x}^{2}} \right)={{e}^{3x}}{{\tan }^{4}}x\times 2{{x}^{2-1}} \\
 & \Rightarrow {{e}^{3x}}{{\tan }^{4}}x\times \dfrac{d}{dx}\left( {{x}^{2}} \right)=2x{{e}^{3x}}{{\tan }^{4}}x...\left( ii \right) \\
\end{align}\]
Now, let us consider the function \[{{x}^{2}}{{\tan }^{4}}x\times \dfrac{d}{dx}\left( {{e}^{3x}} \right)\] . We will have to apply the chain rule which is given by $\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}$ . Let us assume \[y={{e}^{3x}}\] and $u=3x$ . We have to differentiate the second equation with respect to x.
$\dfrac{du}{dx}=3...\left( iii \right)$
Now, we can write y as
\[\Rightarrow y={{e}^{u}}\]
We have to differentiate the above equation with respect to u. We know that $\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}}$ . Therefore, the above equation becomes.
\[\Rightarrow \dfrac{dy}{du}={{e}^{u}}...\left( iv \right)\]
Let us substitute (iii) and (iv) in the chain rule.
$\dfrac{dy}{dx}=3{{e}^{u}}$
We have to substitute back the value of u in the above equation.
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=3{{e}^{3x}} \\
 & \Rightarrow \dfrac{d}{dx}\left( {{e}^{3x}} \right)=3{{e}^{3x}} \\
\end{align}$
Therefore, we can write the value of the second term in the equation (i) as
\[\Rightarrow {{x}^{2}}{{\tan }^{4}}x\times \dfrac{d}{dx}\left( {{e}^{3x}} \right)=3{{x}^{2}}{{e}^{3x}}{{\tan }^{4}}x...\left( v \right)\]
Now, we have to consider \[{{x}^{2}}{{e}^{3x}}\dfrac{d}{dx}\left( {{\tan }^{4}}x \right)\] . We will apply chain rules similarly.
\[\Rightarrow {{x}^{2}}{{e}^{3x}}\dfrac{d}{dx}\left( {{\tan }^{4}}x \right)={{x}^{2}}{{e}^{3x}}\times 4{{\tan }^{3}}x\times \dfrac{d}{dx}\left( \tan x \right)\]
We know that $\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$ .
\[\begin{align}
  & \Rightarrow {{x}^{2}}{{e}^{3x}}\dfrac{d}{dx}\left( {{\tan }^{4}}x \right)={{x}^{2}}{{e}^{3x}}\times 4{{\tan }^{3}}x\times {{\sec }^{2}}x \\
 & \Rightarrow {{x}^{2}}{{e}^{3x}}\dfrac{d}{dx}\left( {{\tan }^{4}}x \right)=4{{x}^{2}}{{e}^{3x}}{{\tan }^{3}}x{{\sec }^{2}}x...\left( vi \right) \\
\end{align}\]
Let us substitute (ii), (v) and (vi) in equation (i)
\[\Rightarrow \dfrac{d}{dx}\left( {{x}^{2}}{{e}^{3x}}{{\tan }^{4}}x \right)=2x{{e}^{3x}}{{\tan }^{4}}x+3{{x}^{2}}{{e}^{3x}}{{\tan }^{4}}x+4{{x}^{2}}{{e}^{3x}}{{\tan }^{3}}x{{\sec }^{2}}x\]
Hence, the derivative of ${{x}^{2}}{{e}^{3x}}{{\tan }^{4}}x$ is \[2x{{e}^{3x}}{{\tan }^{4}}x+3{{x}^{2}}{{e}^{3x}}{{\tan }^{4}}x+4{{x}^{2}}{{e}^{3x}}{{\tan }^{3}}x{{\sec }^{2}}x\] .

Note: Students must be through with all the results of the derivatives of basic functions, properties and rules. We can apply chain rule directly without using the formula of the same. For example, to find the derivative of \[{{\tan }^{4}}x\] , we will consider \[{{\tan }^{4}}x\] as $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ . In this formula, we will substitute x as $\tan x$ . Then, we will find the derivative of $\tan x$ . The product of the result of the last two steps will be the derivative of \[{{\tan }^{4}}x\] .