What is the derivative of #f(x)=arctan(2^x)#?

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Answer 1 · 2018-04-11T14:01:38

#(2^xln2)/(2^(2x)+1)#

Apply the chain rule:

#dy/dx=dy/(du)*(du)/dx#

Let #u=2^x,:.(du)/dx=2^x*ln2#

Then, #y=arctanu,:.dy/(du)=1/(u^2+1)#

And so,

#dy/dx=1/(u^2+1)*2^x*ln2#

#=(2^xln2)/(u^2+1)#

Undo the substitution to get:

#=(2^xln2)/((2^x)^2+1)#

#=(2^xln2)/(2^(2x)+1)#