`h(x) = ln(2x^2 + 1)` Find the derivative of the function.

`h(x) = ln (2x^2+1)`

First, apply the formula:

`(ln u)' = 1/u * u'`

So, the derivative of the function will be:

`h'(x) =1/(2x^2+1)*(2x^2+1)'`

To take the derivative of the inner function, apply the formula:

`(x^n)' = n* x^(n-1)`

`(c)' = 0`

So h'(x) will become:

`h'(x) =1/(2x^2+1) * (2*2x + 0)`

`h'(x) = 1/(2x^2+1) * 4x`

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

Therefore, the derivative of the given function is `h'(x)=(4x)/(2x^2+1)` .