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How do you differentiate $f\left( x \right)={{e}^{x}}\ln x$?

Last updated date: 05th Aug 2024
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Hint: We first define the multiplication rule and how the differentiation of function works. We take addition of these two different differentiated values. We take the $\dfrac{dy}{dx}$ altogether. We keep one function and differentiate the other one and then do the same thing with the other function. Then we take the addition to complete the formula.

Complete step-by-step solution:
We now discuss the multiplication process of two functions where $f\left( x \right)=u\left( x \right)v\left( x \right)$
Differentiating $f\left( x \right)=uv$, we get $\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ uv \right]=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$.
The above-mentioned rule is the multiplication rule. We apply that on $f\left( x \right)={{e}^{x}}\ln x$. We assume the functions where $u\left( x \right)={{e}^{x}},v\left( x \right)=\ln x$
We know that differentiation of $u\left( x \right)={{e}^{x}}$ is ${{u}^{'}}\left( x \right)={{e}^{x}}$ and differentiation of $v\left( x \right)=\ln x$ is ${{v}^{'}}\left( x \right)=\dfrac{1}{x}$. We now take differentiation on both parts of $f\left( x \right)={{e}^{x}}\ln x$ and get $\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ {{e}^{x}}\ln x \right]$.
We place the values of ${{u}^{'}}\left( x \right)={{e}^{x}}$ and ${{v}^{'}}\left( x \right)=\dfrac{1}{x}$ to get
$\dfrac{d}{dx}\left[ {{e}^{x}}\ln x \right]={{e}^{x}}\dfrac{d}{dx}\left( \ln x \right)+\left( \ln x \right)\dfrac{d}{dx}\left( {{e}^{x}} \right)$.
We take all the $\dfrac{dy}{dx}$ forms altogether to get
\begin{align} & \dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ {{e}^{x}}\ln x \right] \\ & \Rightarrow {{f}^{'}}\left( x \right)={{e}^{x}}\times \dfrac{1}{x}+\left( \ln x \right)\left( {{e}^{x}} \right)=\dfrac{{{e}^{x}}}{x}+{{e}^{x}}\ln x \\ & \Rightarrow {{f}^{'}}\left( x \right)={{e}^{x}}\left( \ln x+\dfrac{1}{x} \right) \\ \end{align}
Therefore, differentiation of $f\left( x \right)={{e}^{x}}\ln x$ is ${{e}^{x}}\left( \ln x+\dfrac{1}{x} \right)$.

Note: We need to remember that in the chain rule $\dfrac{d}{d\left[ h\left( x \right) \right]}\left[ goh\left( x \right) \right]\times \dfrac{d\left[ h\left( x \right) \right]}{dx}$, we aren’t cancelling out the part $d\left[ h\left( x \right) \right]$. Cancelation of the base differentiation is never possible. It’s just a notation to understand the function which is used as a base to differentiate. The rule may be extended or generalized to many other situations, including to products of multiple functions, to a rule for higher-order derivatives of a product, and to other contexts.