How do you find the derivative of #arctan(e^x)#?
2 Answers
# d/dx arctan(e^x)= (e^x)/(e^(2x)+1) #
Explanation:
When tackling the derivative of inverse trig functions. I prefer to rearrange and use Implicit differentiation as I always get the inverse derivatives muddled up, and this way I do not need to remember the inverse derivatives. If you can remember the inverse derivatives then you can use the chain rule.
#y=arctan(e^x) <=> tany=e^x #
Differentiate Implicitly:
# sec^2ydy/dx = e^x # ..... [1]
Using the
# tan^2y+1 -= sec^2y #
# :. (e^x)^2+1=sec^2y #
# :. e^(2x)+1=sec^2y #
Substituting into [1]
# :. (e^(2x)+1)dy/dx=e^x #
# :. dy/dx = (e^x)/(e^(2x)+1) #
Explanation:
Using implicit differentiation together with the known derivatives
let
If we draw a right triangle with an angle
#=e^x/sec^2(y)#
#=e^x/(e^(2x)+1)#