# How do you differentiate  f(x)=e^((6x-2)^2  using the chain rule.?

Apr 6, 2016

$h ' \left(x\right) = 12 \left(6 x - 2\right) \cdot {e}^{{\left(6 x - 2\right)}^{2}} = \left(72 x - 24\right) \cdot {e}^{{\left(6 x - 2\right)}^{2}}$

#### Explanation:

Given : f(x)=e^((6x-2)^2

Required: The derivative of $f \left(x\right)$

Definition and principles: The chain rule
Let $h \left(x\right) = \left(f \circ g\right) \left(x\right)$ then the derivative of h(x) is

$h ' \left(x\right) = f ' \left(g \left(x\right)\right) g ' \left(x\right)$

Solution Strategy: let $f \left(x\right) = {\left(6 x - 2\right)}^{2}$ and $g \left(x\right) = {e}^{f \left(x\right)}$
A) Evaluate $f ' \left(g \left(x\right)\right) \mathmr{and}$g'(x) B) Put it all together - f'(g(x))*g'(x) 

A) I am going to write the chain rule as:
let $g \left(x\right) = {\left(6 x - 2\right)}^{2}$, then $g ' \left(x\right) = 12 \left(6 x - 2\right)$
now $f \left(g \left(x\right)\right) = {e}^{g \left(x\right)}$ and f'(g(x)) = e^g(x)=color(red)(e^((6x-2)^2)
B) f'(g(x))*g'(x) = color(red)(e^((6x-2)^2)*color(blue)(12(6x-2))#

$h ' \left(x\right) = 12 \left(6 x - 2\right) \cdot {e}^{{\left(6 x - 2\right)}^{2}} = \left(72 x - 24\right) \cdot {e}^{{\left(6 x - 2\right)}^{2}}$
$\frac{\mathrm{dh} \left(x\right)}{\mathrm{dx}} = \frac{\mathrm{df} \left(u\right)}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$ and you let
$u = {\left(6 x - 2\right)}^{2}$ and $f \left(u\right) = {e}^{u}$ then
$\frac{\mathrm{df} \left(u\right)}{\mathrm{du}} = {e}^{u} \mathmr{and} \frac{\mathrm{du}}{\mathrm{dx}} = 12 \left(6 x - 2\right)$
and $\frac{\mathrm{dh} \left(x\right)}{\mathrm{dx}} = {\left[{e}^{u} \cdot 12 \left(6 x - 2\right)\right]}_{u = {\left(6 x - 2\right)}^{2}} = \left(72 x - 24\right) \cdot {e}^{{\left(6 x - 2\right)}^{2}}$