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

Thank you!

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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