How do you find the second derivative of $h(x)=ln(2x^2+1)$ ?

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Answer 1 · 2016-07-17T19:41:55

$=(4-8x^2)/((2x^2+1)^2)$

We use the chain rule to compute the derivative.

$(dh)/(dx) = (dh)/(du)(du)/(dx)$

$1/(2x^2+1)*4x = (4x)/(2x^2+1)$

Now we need to use the quotient rule.

$(d^2h)/(dx^2) = (4(2x^2+1) - 16x^2)/((2x^2+1)^2)$

$=(4-8x^2)/((2x^2+1)^2)$

How do you find the second derivative of h(x)=ln(2x^2+1)h(x)=ln(2x^2+1) ?