#d/dx(tan^(-1)2x)=2/(4x^2+1)#. Why is there a #2# in the numerator?

Question

d/dx(tan^(-1)2x) = 2/(4x^2+1). Why is there a #2# in the numerator?

I know how to get to the denominator but not the numerator. Can anyone just briefly explain? Nothing lengthy needed.

Thanks!

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Answer 1 · 2017-12-18T12:29:21

The factor of #2# occurs as a results of applying the chain rule.

We have a known result:

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

Then applying the chain rule we have

# d/dx tan^(-1)2x =1/(1+(2x)^2) d/dx (2x)#
# " " =2/(1+4x^2) #

If that was unclear we can perform an explicit substitution, viz:

Let #{ (y=tan^(-1)2x), (u=2x):} => { (y=tan^(-1)u,), ((du)/dx=2,dy/(du)=1/(1+u^2)) :} #

Then by the chain rule we have:

# dy/dx = (dy)/(du) \ (du)/dx #
# \ \ \ \ \ \ = 1/(1+u^2) \ 2 #
# \ \ \ \ \ \ = 2/(1+u^2) #

And restoring the substitution we get the result:

# dy/dx = 2/(1+(2u)^2) = 2/(1+4u^2) #