Question

Differentiate $x{e}^x$ From First Principles.

Answer 1

Step 1: Find the derivative of $f\left(x\right)=x{e}^x$

Given,$x{e}^x$

Let , $f\left(x\right)=x{e}^x$

Derivative of a function $f\left(x\right)$ is given by:

$f^{\mathrm{\prime }}\left(x\right)=\underset{h\to 0}{lim} \frac{f(x+h)-f\left(x\right)}{h}$

{where $h$ is a very small positive number}

The derivative of $f\left(x\right)=x{e}^x$ is given as:

$\begin{array}{c}{f}^{\mathrm{\prime }}\left(x\right)=\underset{h\to 0}{lim} \frac{f(x+h)-f\left(x\right)}{h}\\ f^{\mathrm{\prime }}\left(x\right)=\underset{h\to 0}{lim} \frac{(x+h)e^{(x+h)}-x{e}^x}{h}\end{array}$

Now by using the algebra of limits:

$\begin{array}{c}\Rightarrow f^{\mathrm{\prime }}\left(x\right)=\underset{h\to 0}{lim} \frac{h{e}^{x+h}}{h}+\underset{h\to 0}{lim} \frac{x\left(e^{x+h}-e^x\right)}{h}\\ \Rightarrow f^{\mathrm{\prime }}\left(x\right)=\underset{h\to 0}{lim} e^{x+h}+\underset{h\to 0}{lim} \frac{x{e}^x\left(e^h-1\right)}{h}\end{array}$

Step 2: Again by using the algebra of limits:

$\Rightarrow f^{\mathrm{\prime }}\left(x\right)=e^{x+0}+\underset{h\to 0}{lim} \frac{e^h-1}{h}\times \underset{h\to 0}{lim} x{e}^x$

Now by using the formula:

$\begin{array}{c}\underset{x\to 0}{lim} \frac{e^x-1}{x}={\log }_{e}e=1\\ f^{\mathrm{\prime }}\left(x\right)=e^x+x{e}^x\\ f^{\mathrm{\prime }}\left(x\right)=e^x(x+1)\end{array}$

Therefore, the derivation of $f\left(x\right)=x{e}^x=e^x(x+1)$.