Question

How do you find the derivative of the function $y={{e}^{x}}.\ln x$?

Answer 1

We start solving the problem by applying the differentiation on both sides of the given equation with respect to x. We then make use of the uv rule of differentiation $\dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ to proceed through the problem. We then make use of the results $\dfrac{d\left( \ln x \right)}{dx}=\dfrac{1}{x}$, $\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}}$ to proceed further through the problem. We then make the necessary calculations to get the required derivative of the given function.

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

$\therefore$ The derivative of the given function $y={{e}^{x}}.\ln x$ as $\dfrac{dy}{dx}=\dfrac{{{e}^{x}}\left( 1+x\ln x \right)}{x}$.

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