Question #e3a4a

2 Answers
Oct 23, 2017

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}}

Oct 23, 2017

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))