#### Provide solution RD Sharma maths class 12 chapter  10 differentiability exercise 10.1 question 8 maths textbook solution

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

Hint:

Use first principle formula to find the differentiation

Given:

${x^{2}}e^{x}$

Solution:

Let,

\begin{aligned} &f(x)=x^{2} e^{x} \\ &f(x+h)=(x+h)^{2} \cdot e^{(x+h)} \end{aligned}

Now, we will use the formula of first principle

\begin{aligned} &\frac{d}{d x}(f(x))=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &\frac{d}{d x}\left(x^{2} e^{x}\right)=\lim _{h \rightarrow 0} \frac{(x+h)^{2} \cdot e^{(x+h)}-x^{2} e^{x}}{h} \end{aligned}

$=\lim _{h \rightarrow 0} \frac{\left(x^{2}+2 h x+h^{2}\right) e^{x+h}-x^{2} e^{x}}{h}$                                    $\left [ \because \left ( a+b \right )^{2}= a^{2}+2ab+b^{2} \right ]$

\begin{aligned} &=\lim _{h \rightarrow 0}\left[\frac{\left(x^{2} e^{x+h}-x^{2} e^{x}\right)}{h}+\frac{2 h x e^{x+h}}{h}+\frac{h^{2} e^{x+h}}{h}\right] \\ &=\lim _{h \rightarrow 0}\left[\frac{x^{2} e^{x} \cdot e^{h}-x^{2} e^{x}}{h}+2 x e^{x+h}+h e^{x+h}\right] \\ &=\lim _{h \rightarrow 0}\left[\frac{x^{2} e^{x}\left(e^{h}-1\right)}{h}+2 x e^{x+h}+h e^{x+h}\right] \\ &=\lim _{h \rightarrow 0}\left[x^{2} e^{x} \cdot\left(\frac{e^{h}-1}{h}\right)+2 x e^{x+h}+h e^{x+h}\right] \end{aligned}

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

\begin{aligned} &=x^{2} e^{x}+2 x e^{x}+\left(0 \times e^{x}\right) \\ &=x^{2} e^{x}+2 x e^{x} \\ &=e^{x}\left(x^{2}+2 x\right) \\ &\therefore \frac{d}{d x}\left(x^{2} e^{x}\right)=e^{x}\left(x^{2}+2 x\right) \end{aligned}

Hence, the differentiation of $x^{2} e^{x}$ is $e^{x}\left ( x^{2} +2x \right )$