# Question e3a4a

Oct 23, 2017

Look below

#### Explanation:

since its using e, the format should be ${e}^{\sqrt{\tan} x}$

let $f \left(x\right) = {e}^{x}$
and let $g \left(x\right) = \sqrt{\tan} x$

the chain rule is as stated $f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

${e}^{\sqrt{\tan} x} \cdot g ' \left(x\right)$

now find the derivative of $\sqrt{\tan} x$ by using the chain rule

f(x) = $\sqrt{x}$
g(x) = $\tan x$

the derivative of $\sqrt{x}$ is $- \frac{1}{2 {x}^{\frac{3}{2}}}$

the derivative of $\tan x$ is ${\sec}^{2} \left(x\right)$

now use chain rule

$- \frac{1}{2 {x}^{\frac{3}{2}} \tan x} \cdot {\sec}^{2} \left(x\right)$

$- {\sec}^{x} / \left\{2 \tan {x}^{\frac{1}{2}}\right\}$

Oct 23, 2017

f’(x) = (e^(sqrt(tan x)) * sec^2 x) / (2 sqrt ( tan x))

#### Explanation:

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

f’(x) = e^(sqrt(tan x)) * ( sqrt(tans) dx) 

f’(x) = e^(sqrt(tans)) * (1/(2 sqrt(tan x)) )* (tan x dx)

f’(x) = e^(sqrt(tans)) * (1/(2 sqrt(tan x)) )* sec ^x

f’(x) = (e^(sqrt(tan x)) * sec^2 x) / (2 sqrt ( tan x))#