Question

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

Answer 1

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

Explanation:

First, let's rewrite this function.
`1/(2x^2+1)=>(2x^2+1x^0)^-1`

Now, remember the power rule and the chain rule.

Power rule: `d/dx(x^n)=nx^(n-1)` where `n` is a constant.

Chain rule: If `f(x)=g(h(x))`, then `f'(x)=g'(h(x))*h'(x)`

Therefore, we have:

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

=>`-(2x^2+1)^-2*4x^(2-1)+0x^-1`

=>`-(2x^2+1)^-2*4x^(1)+0`

=>`-1/(2x^2+1)^2*4x`

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

That is our answer!