# Find derivative? f(x) = (xe^x)^2

Jun 17, 2018

${f}^{'} \left(x\right) = 2 x {e}^{2 x} \left(x + 1\right)$

#### Explanation:

Differentiation of product of two functions:

$\frac{d}{d x} \left(f \left(x\right) g \left(x\right)\right) = {f}^{'} \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

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

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

$\therefore {f}^{'} \left(x\right) = 2 x {e}^{2 x} \left(x + 1\right)$ [Ans]

Jun 17, 2018

$f ' \left(x\right) = 2 x \left(x + 1\right) {e}^{2 x}$

#### Explanation:

Here,

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

Let,

$y = {u}^{2} , w h e r e , \textcolor{red}{u = x \cdot {e}^{x}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{du}} = 2 u \ldots \ldots . \to \left(1\right)$ .

$\text{ Using "color(blue)"Product Rule}$ , we get

$\frac{\mathrm{du}}{\mathrm{dx}} = x \cdot \frac{d}{\mathrm{dx}} \left({e}^{x}\right) + {e}^{x} \cdot \frac{d}{\mathrm{dx}} \left(x\right)$

$\implies \frac{\mathrm{du}}{\mathrm{dx}} = x \cdot {e}^{x} + {e}^{x} \cdot 1. . . \to \left(2\right)$

Now ,$\text{using "color(blue)"Chain Rule} :$

color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dx)

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 \textcolor{red}{u}\right) \times \left(x {e}^{x} + {e}^{x}\right) \ldots . \to F r o m \left(1\right) \mathmr{and} \left(2\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \textcolor{red}{\left(x {e}^{x}\right)} \times \left(x {e}^{x} + {e}^{x}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {e}^{x} \times {e}^{x} \left(x + 1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x \left(x + 1\right) {e}^{2 x}$
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OR

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

$\implies f ' \left(x\right) = {x}^{2} \frac{d}{\mathrm{dx}} \left({e}^{2 x}\right) + {e}^{2 x} \frac{d}{\mathrm{dx}} \left({x}^{2}\right)$

$\implies f ' \left(x\right) = {x}^{2} {e}^{2 x} \cdot 2 + {e}^{2 x} \left(2 x\right) = 2 {x}^{2} {e}^{2 x} + 2 x {e}^{2 x}$

Hence,

$f ' \left(x\right) = \left(2 {x}^{2} + 2 x\right) {e}^{2 x} = 2 x \left(x + 1\right) {e}^{2 x}$