Question

Differentiate from first principle:

$\left(iv\right)x{e}^x$

Answer 1

Given:

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

The derivative of a function $f\left(x\right)$ is defined as:

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

Putting $f\left(x\right)$ in the above expression, we get:

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

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

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

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

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

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

$\Rightarrow f^{\prime }\left(x\right)=x{e}^x+e^x.1$

$\Rightarrow f^{\prime }\left(x\right)=(x+1)e^x$

Hence, the derivative of $x{e}^x$ is $(x+1)e^x$