# How do you differentiate g(x) = sqrt(x-3)e^(4x)cos(7x) using the product rule?

Mar 14, 2018

$\frac{d}{\mathrm{dx}} f \left(x\right) = \frac{1 - 12 {e}^{4 x}}{2 \sqrt{x - 3 {e}^{4 x}}} \left(\cos \left(7 x\right)\right) + \left(- 7 \sin \left(7 x\right)\right) \left(\sqrt{x - 3 {e}^{4 x}}\right)$

#### Explanation:

I'm assuming your original post meant $\sqrt{x - 3 {e}^{4 x}} \cos \left(7 x\right)$ and this was a simple typo in inferring this question. Therefore let us begin by recalling the product rule.

$\left(f g\right) ' = f ' g + f g '$ where $f \mathmr{and} g$ are differentiable.

Let us define $f \mathmr{and} g$

$f \left(x\right) = \sqrt{x - 3 {e}^{4 x}}$
$g \left(x\right) = \cos \left(7 x\right)$

To simplify this problem into bite size parts let us differentiate each function separately and plug in our response at the end of the problem.

Therefore,
$\frac{d}{\mathrm{dx}} f \left(x\right)$ requires us to use the chain rule.

where our inner function is ${x}^{\frac{1}{2}}$ and our outer function is $\left(x - 3 {e}^{4 x}\right)$

recall $\left(f \circ g\right) ' = \left(f ' \circ g\right) \cdot g '$

$f ' = \left(\frac{1}{2}\right) \left({x}^{- \frac{1}{2}}\right)$ or
$\frac{1}{2 \sqrt{x}}$ of our inner function where we must plug in $g = \left(x - 3 {e}^{4 x}\right)$
(1)/(2sqrt((x-3e^(4x))) now find $g '$

$g = \left(x - 3 {e}^{4 x}\right)$

$g ' = 1 - 12 {e}^{4 x}$ recalling that differentiation splits across sums and differences making two derivatives. Noting $\frac{d}{\mathrm{dx}} {e}^{x} = \left(x '\right) {e}^{x}$

Where our $x = \left(4 x\right)$

Furthermore,
d/dx f(x) = (1-12e^(4x))/(2sqrt(x-3e^(4x))

Now let us find $\frac{d}{\mathrm{dx}} g \left(x\right) = \cos \left(7 x\right)$ recalling that we must use the chain rule again.

$\frac{d}{\mathrm{dx}} g \left(x\right) = - 7 \sin \left(7 x\right)$ where

$f = \cos \left(x\right) \mathmr{and} g = 7 x$
$\left(f \circ g\right) ' = \left(f ' \circ g\right) \cdot g '$

$\frac{d}{\mathrm{dx}} \cos \left(x\right) = - \sin \left(x\right)$

$\frac{d}{\mathrm{dx}} 7 x = 7$

Recall our original problem $\sqrt{x - 3 {e}^{4 x}} \cos \left(7 x\right)$
$\left(f g\right) ' = f ' g + f g '$ where $f \mathmr{and} g$ are differentiable.

$f = \sqrt{x - 3 {e}^{4 x}} \mathmr{and} g = \cos \left(7 x\right)$ also

f' = (1-12e^(4x))/(2sqrt(x-3e^(4x)) and $g ' = - 7 \sin \left(7 x\right)$

$\frac{1 - 12 {e}^{4 x}}{2 \sqrt{x - 3 {e}^{4 x}}} \left(\cos \left(7 x\right)\right) + \left(- 7 \sin \left(7 x\right)\right) \left(\sqrt{x - 3 {e}^{4 x}}\right)$