How do you find the derivative of #arctan(e^x)#?

2 Answers
Steve M
Dec 22, 2016

# 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"/"sec# identity;

# 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) #

sente
Dec 22, 2016

#d/dxarctan(e^x)=e^x/(e^(2x)+1)#

Explanation:

Using implicit differentiation together with the known derivatives #d/dx tan(x) = sec^2(x)# and #d/dx e^x = e^x#,

let #y = arctan(e^x)#

#=> tan(y) = tan(arctan(e^x)) = e^x#

#=> d/dxtan(y) = d/dxe^x#

#=> sec^2(y)dy/dx = e^x#

#=> dy/dx = e^x/sec^2(y)#

If we draw a right triangle with an angle #y# and legs such that #tan(y) = e^x#, then we find that #sec(y) = sqrt(e^(2x)+1)#. Using that, we get our final result:

#d/dxarctan(e^x) = dy/dx#

#=e^x/sec^2(y)#

#=e^x/(e^(2x)+1)#

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