$\displaystyle y = (x^3 - 2x)^4$
$\displaystyle y' = 4(x^3 - 2x)^3 (3x^2 - 2)$
$\displaystyle y'' = ?$
I'm having difficulty proceeding from there to get the second derivative. The answer is $\displaystyle 12(x^3 - 2x)^2 (11x^4 - 16x^2 + 4)$
$\displaystyle y = (x^3 - 2x)^4$
$\displaystyle y' = 4(x^3 - 2x)^3 (3x^2 - 2)$
$\displaystyle y'' = ?$
I'm having difficulty proceeding from there to get the second derivative. The answer is $\displaystyle 12(x^3 - 2x)^2 (11x^4 - 16x^2 + 4)$
Yes, now you can take out common factors
$\displaystyle 4\left\{(x^3 - 2x)^3 \frac{d}{dx}(3x^2 - 2) + (3x^2 - 2)\frac{d}{dx}[(x^3 - 2x)^3]\right\}$
$\displaystyle = 4[6x(x^3 - 2x)^3 + 3(3x^2 - 2)(3x^2 - 2)(x^3 - 2x)^2]$
Now take out common factors.
$\displaystyle = 12(x^3 - 2x)^2[2x(x^3 - 2x) + (3x^2 - 2)^2]$
$\displaystyle = 12(x^3 - 2x)^2[2x^4 - 4x^2 + 9x^4 - 12x^2 + 4]$
$\displaystyle = 12(x^3 - 2x)^2(11x^4 - 16x^2 + 4)$