# Find the second derivative of the function. f(x) = 2(x² - 1)^3π = pi I managed to figure out the first derivative, but finding the second one is difficult

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`f(x)=2(x^2-1)^3`

`f'(x)=2*3(x^2-1)^2(2x)=12x(x^2-1)^2`

`f''(x)=12(x^2-1)^2+12x(2)(x^2-1)(2x)`

`=12(x^2-1)^2+48x^2(x^2-1)`

`=60x^4-72x^2+12`

As a check, expand the binomial and then take the derivative -- bypassing the product and chain rules:

`f(x)=2(x^2-1)^3=2[x^3-3x^4+3x^2-1]=2x^6-6x^4+6x^2-2`

`f'(x)=12x^5-24x^3+12x`

`f''(x)=60x^4-72x^2+12`