Question

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

Answer 1

Use the rule `(f/g)´=(f´·g-f·g´)/g^2`

Explanation:

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

Then

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

Where `f=x` and `g=1+x^2`

Answer 2

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

Explanation:

`"differentiate using the "color(blue)"quotient rule"`

`"given "f(x)=(g(x))/(h(x))" then"`

`f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larrcolor(blue)"quotient rule"`

`g(x)=xrArrg'(x)=1`

`h(x)=1+x^2rArrh'(x)=2x`

`d/dx(x/(1+x^2))`

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