Question

How do you find the derivative of `f(x)= 2/(1+x^2)`?

Answer 1

`y=f(x)`

`y=2/(1+x^2)`

`y^-1=(1+x^2)/2`

`2y^-1=1+x^2`

Now use implicit differentiation...

`-2y^-2*(dy)/(dx)=2x`

`-2/(y^2)*(dy)/(dx)=2x`

Divide expressions on both sides of the equation by 2...

`-1/y^2*(dy)/(dx)=x`

Multiply expressions on both sides of the equation by -1...

`1/y^2*(dy)/(dx)=-x`

Now multiply expressions on both sides of the equation by y^2...

`(dy)/(dx)=-x*y^2`

Don't forget the real value of y...

`(dy)/(dx)=-x*(2/(1+x^2))^2`

Clean up the final result...

`(dy)/(dx)=-(4x)/((1+x^2)^2)`

Which means that...

`f'(x)=-(4x)/((1+x^2)^2)`