Calculus/Calc I

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Question
I need help and an explanation of how to find the derivative of y=e^tanx all under a square root

Thank you!

Answer
y =sqrt (e^tanx)  =  (e^tanx)^(1/2)

Differentiating both sides wrt x,

dy/dx = (1/2)(sec ^2  x) (e^tanx)^(-1/2)
      = (sec ^2  x)/[ 2* sqrt (e^tanx)]  (shown)

Note that we employ the understanding via the chain rule effect that differentiating y=[e^f(x)]^n  with respect to x gives  dy/dx = n* f'(x) * [e^f(x)]^(n-1) , where f(x)=tan x,  f'(x)= sec ^2  x  and n = 1/2 for the above.


Hope this helps. Peace.

Calculus

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