# What is the derivative of (x^2)(e^(4x))?

Jan 28, 2016

$2 x {e}^{4 x} \left(2 x + 1\right)$

#### Explanation:

Differentiate $\textcolor{b l u e}{\text{ using product rule }}$

$\frac{d}{\mathrm{dx}} \left({x}^{2} {e}^{4 x}\right) = {x}^{2} \frac{d}{\mathrm{dx}} \left({e}^{4 x}\right) + {e}^{4 x} \frac{d}{\mathrm{dx}} \left({x}^{2}\right)$

$= {x}^{2} \left({e}^{4 x} \frac{d}{\mathrm{dx}} \left(4 x\right)\right) + {e}^{4 x} \left(2 x\right)$

$= {x}^{2.} {e}^{4 x} .4 + {e}^{4 x} .2 x$

$= 4 {x}^{2} {e}^{4 x} + 2 x {e}^{4 x} = 2 x {e}^{4 x} \left(2 x + 1\right)$

Jan 28, 2016

$2 x {e}^{4 x} \left(2 x + 1\right)$

#### Explanation:

suppose,
$f \left(x\right) = {x}^{2} {e}^{4 x}$

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right)$

$= \frac{d}{\mathrm{dx}} \left({x}^{2} {e}^{4 x}\right)$

$= {x}^{2} \frac{d}{\mathrm{dx}} \left({e}^{4 x}\right) + {e}^{4 x} \frac{d}{\mathrm{dx}} \left({x}^{2}\right)$

$= {x}^{2} {e}^{4 x} \frac{d}{\mathrm{dx}} \left(4 x\right) + 2 x {e}^{4 x}$

$= 4 {x}^{2} {e}^{4 x} + 2 x {e}^{4 x}$

$= 2 x {e}^{4 x} \left(2 x + 1\right)$

Jan 28, 2016

$2 x {e}^{4 x} \left(2 x + 1\right)$

#### Explanation:

Use the product rule, which states that for a function $y = f \left(x\right) g \left(x\right)$,

$y ' = f ' \left(x\right) g \left(x\right) + g ' \left(x\right) f \left(x\right)$

This can be rewritten in English as: when two functions are multiplied by one another, their derivative is the derivative of one times the nondifferentiated version of the other, plus the derivative of the other function times the function that was differentiated in the other term.

Here, we have

$f \left(x\right) = {x}^{2}$
$g \left(x\right) = {e}^{4 x}$

So, their derivatives are

$f ' \left(x\right) = 2 x$
$g ' \left(x\right) = 4 {e}^{4 x}$

Note that finding $g ' \left(x\right)$ require the chain rule. The chain rule states that for a function $y = h \left(k \left(x\right)\right)$,

$y ' = h ' \left(k \left(x\right)\right) \cdot k ' \left(x\right)$

In the case of ${e}^{4 x}$,

$h \left(x\right) = {e}^{x}$
$k \left(x\right) = 4 x$

$h ' \left(x\right) = {e}^{x}$
$k ' \left(x\right) = 4$

So, plugging these into the chain rule, we see that

$\frac{d}{\mathrm{dx}} \left({e}^{4 x}\right) = {e}^{4 x} \cdot 4 = 4 {e}^{4 x}$

Returning to the previous set of functions and derivatives, we use the product rule to see that

$\frac{d}{\mathrm{dx}} \left({x}^{2} {e}^{4 x}\right) = 2 x {e}^{4 x} + 4 {x}^{2} {e}^{4 x}$

This can be simplified as

$= 2 x {e}^{4 x} \left(2 x + 1\right)$