$\displaystyle y = (x^3 - 2x)^4$

$\displaystyle y' = 4(x^3 - 2x)^3 (3x^2 - 2)$

$\displaystyle y'' = ?$

I'm having difficulty proceeding from there to get the second derivative. The answer is $\displaystyle 12(x^3 - 2x)^2 (11x^4 - 16x^2 + 4)$

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- Nov 13th 2009, 06:34 PMArchduke01Finding the second derivative of a function
$\displaystyle y = (x^3 - 2x)^4$

$\displaystyle y' = 4(x^3 - 2x)^3 (3x^2 - 2)$

$\displaystyle y'' = ?$

I'm having difficulty proceeding from there to get the second derivative. The answer is $\displaystyle 12(x^3 - 2x)^2 (11x^4 - 16x^2 + 4)$ - Nov 13th 2009, 06:37 PMProve It
- Nov 13th 2009, 06:44 PMArchduke01
- Nov 13th 2009, 06:56 PMProve It
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.

$\displaystyle = 12(x^3 - 2x)^2[2x(x^3 - 2x) + (3x^2 - 2)^2]$

$\displaystyle = 12(x^3 - 2x)^2[2x^4 - 4x^2 + 9x^4 - 12x^2 + 4]$

$\displaystyle = 12(x^3 - 2x)^2(11x^4 - 16x^2 + 4)$ - Nov 13th 2009, 07:08 PMArchduke01
- Nov 13th 2009, 07:10 PMProve It
- Nov 13th 2009, 07:12 PMArchduke01
- Nov 13th 2009, 07:15 PMProve It
- Nov 13th 2009, 07:17 PMArchduke01