Question #e3a4a

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Answer 1 · 2017-10-23T19:26:40

Look below

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 · 2017-10-23T19:55:30

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

#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))#