The given function: `f(x) =arctan(e^x)` is in a form of inverse trigonometric function.
It can be evaluated using the derivative formula for inverse of tangent function:
`d/(dx)arctan(u) = ((du)/(dx))/(1+x^2)` .
We let `u = e^x` then `(du)/(dx)= d/(dx) (e^x)= e^x` .
Applying the the formula, we get:
`f'(x)= d/(dx) arctan(e^x)`
`=e^x/(1 +(e^x)^2)`
Using the law of exponent: `(x^n)^m=x^(n*m)` , we may simplify the part:
`(e^x)^2 = e^((x*2)) = e^(2x)`
The derivative of the function` f(x) = arctan(e^x)` becomes:
`f'(x)= e^x/(1 +e^(2x)) ` as the Final Answer.
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