Question

Question #e3a4a

Answer 1

Look below

Explanation:

since its using e, the format should be `e^{sqrttanx}`

let `f(x)=e^x`
and let `g(x)=sqrttanx`

the chain rule is as stated `f'(g(x))*g'(x)`

`e^{sqrttanx}* g'(x)`

now find the derivative of `sqrttanx` by using the chain rule

f(x) = `sqrtx`
g(x) = `tanx`

the derivative of `sqrtx` is `-1/{2x^{3/2}}`

the derivative of `tanx` is `sec^2(x)`

now use chain rule

`-1/{2x^{3/2}tanx}*sec^2(x)`

`-sec^x/{2tanx^{1/2}}`

Answer 2

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

Explanation:

`f (x) = 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))`