Differentiate each of the following functions from the first principal:

We have to find the derivative of $x^{2} e^{x}$ with the first principle method, so,

$f(x)=x^{2} e^{x}$

by using the first principle formula, we get,

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{(x+h)^{2} e^{(x+h)}-x^{2} e^{x}}{h}$

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

[By using $\left.(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

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

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{x^{2} e^{x}\left(e^{h}-1\right)}{h}+\lim _{h \rightarrow 0} \frac{e^{(x+h)}\left[h^{2}+2 h x\right]}{h}$

[By using $\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1$ ]

$f^{\prime}(x)=x^{2} e^{x}+\lim _{h \rightarrow 0} e^{(x+h)}[h+2 x]$

$f^{\prime}(x)=x^{2} e^{x}+2 x e^{x}$