Question

What is the derivative of `x/(x^2+1)`?

Answer 1

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

Explanation:

If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:

`d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2`, or less formally, `" "(u/v)' = (v(du)-u(dv))/v^2`

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with `y=(x)/(x^2+1)` Then

`{ ("Let "u=x, => , (du)/dx=1), ("And "v=x^2+1, =>, (dv)/dx=2x ) :}`

`:. d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2`
`" "dy/dx = ( (x^2+1)(1) - (x)(2x) ) / (x^2+1)^2`
`" "dy/dx = ( x^2+1 - 2x^2 ) / (x^2+1)^2`
`" "dy/dx = ( 1-x^2 ) / (x^2+1)^2`