Question

How do you find the derivative of the function $\dfrac{{{e}^{4x}}}{x}$?

Answer 1

Hint: We start solving the problem by equating the given function to a variable. We then make use of the fact that the derivative of the function $\dfrac{u}{v}$ as $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$ to proceed through the problem. We then make use of the facts that $\dfrac{d}{dx}\left( {{e}^{ax}} \right)=a{{e}^{ax}}$ and $\dfrac{d}{dx}\left( x \right)=1$ to proceed further through the problem. We then make the necessary calculations in the obtained result to get the required answer for the given problem.

Complete step by step answer:
According to the problem, we are asked to find the derivative of the function $\dfrac{{{e}^{4x}}}{x}$.
Let us assume $y=\dfrac{{{e}^{4x}}}{x}$ ---(1).
Let us differentiate both sides of equation (1) with respect to x.
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( \dfrac{{{e}^{4x}}}{x} \right)$ ---(2).
We know that the derivative of the function $\dfrac{u}{v}$ is defined as $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$. Let us use this result in equation (2).
$\Rightarrow \dfrac{dy}{dx}=\dfrac{x\dfrac{d}{dx}\left( {{e}^{4x}} \right)-{{e}^{4x}}\dfrac{d}{dx}\left( x \right)}{{{x}^{2}}}$ ---(3).
We know that $\dfrac{d}{dx}\left( {{e}^{ax}} \right)=a{{e}^{ax}}$ and $\dfrac{d}{dx}\left( x \right)=1$. Let us use these results in equation (3).
$\Rightarrow \dfrac{dy}{dx}=\dfrac{x\left( 4{{e}^{4x}} \right)-{{e}^{4x}}\left( 1 \right)}{{{x}^{2}}}$.
$\Rightarrow \dfrac{dy}{dx}=\dfrac{4x{{e}^{4x}}-{{e}^{4x}}}{{{x}^{2}}}$.
$\Rightarrow \dfrac{dy}{dx}=\dfrac{{{e}^{4x}}\left( 4x-1 \right)}{{{x}^{2}}}$.
So, we have found the derivative of the given function $\dfrac{{{e}^{4x}}}{x}$ as $\dfrac{{{e}^{4x}}\left( 4x-1 \right)}{{{x}^{2}}}$.

$\therefore $ The derivative of the given function $\dfrac{{{e}^{4x}}}{x}$ is $\dfrac{{{e}^{4x}}\left( 4x-1 \right)}{{{x}^{2}}}$.

Note: We should perform each step carefully in order to avoid confusion and calculation mistakes while solving this problem. We can also solve the given problem as shown below:
We have $y=\dfrac{{{e}^{4x}}}{x}$.
$\Rightarrow y={{e}^{4x}}\times \dfrac{1}{x}$ ---(4).
Let us differentiate both sides of equation (4) with respect to x.
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{e}^{4x}}\times \dfrac{1}{x} \right)$ ---(5).
We know that the derivative of the function $uv$ is defined as $\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$. Let us use this result in equation (5).
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{x}\dfrac{d}{dx}\left( {{e}^{4x}} \right)-{{e}^{4x}}\dfrac{d}{dx}\left( \dfrac{1}{x} \right)$ ---(6).
We know that $\dfrac{d}{dx}\left( {{e}^{ax}} \right)=a{{e}^{ax}}$ and $\dfrac{d}{dx}\left( \dfrac{1}{x} \right)=\dfrac{-1}{{{x}^{2}}}$. Let us use these results in equation (3).
$\Rightarrow \dfrac{dy}{dx}={{e}^{4x}}\left( \dfrac{-1}{{{x}^{2}}} \right)+\dfrac{1}{x}\left( 4{{e}^{4x}} \right)$.
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-{{e}^{4x}}}{{{x}^{2}}}+\dfrac{4{{e}^{4x}}}{x}\].
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{4x{{e}^{4x}}-{{e}^{4x}}}{{{x}^{2}}}\].
$\Rightarrow \dfrac{dy}{dx}=\dfrac{{{e}^{4x}}\left( 4x-1 \right)}{{{x}^{2}}}$.