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)#