What is the derivative of #y= Ln tan^-1 x#?

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Answer 1 · 2018-05-28T14:24:47

#1/((tan^-1x)(x^2+1))#

Given: #y=ln(tan^-1x)#.

Use the chain rule, which states that,

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

Let #u=tan^-1x,:.(du)/dx=1/(x^2+1)#.

Then, #y=lnu,:.dy/(du)=1/u#.

Combining, we get:

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

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

Replacing back #u=tan^-1x#, we get:

#=1/((tan^-1x)(x^2+1))#