Question

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

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!

Answer 1

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)`