Question

Differentiate from first principle:

$\left(vii\right)x^2{e}^x$

Answer 1

Given:

$f\left(x\right)=x^2{e}^x$

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 above expression, we get:

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{(x+h{)}^2{e}^{x+h}-x^2{e}^x}{h}$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{(x^2+2xh+h^2)e^x{e}^h-x^2{e}^x}{h}$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{x^2{e}^x(e^h-1)+(2xh+h^2)e^x{e}^h}{h}$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{x^2{e}^x(e^h-1)}{h}+\underset{h\to 0}{lim}(2x+h)e^x{e}^h$

$\Rightarrow f^{\prime }\left(x\right)=x^2{e}^x+(2x+0)e^x{e}^0(\because \underset{h\to 0}{lim}\frac{e^h-1}{h}=1)$
$\Rightarrow f^{\prime }\left(x\right)=(x^2+2x).e^x$

Therefore, the derivative of $x^2{e}^x$ is $(x^2+2x)e^x$