To solve something like:
$\displaystyle f(x)= (8x^2 + 2)^6 * (3x^2-9)^(14) $
Would I use the chain rule twice?
I enter in:
$\displaystyle (48x^2+12)^5 *(3x^2-9)^{14} + (8x^2+2)^6 * (42x^2-126)^{13}$
And it comes back as incorrect, what do I need to do next?
Wait that's not my question.
I now want to know what to do with the 16x
in
$\displaystyle f'(x)= 6(8x^2+2)^5*(16x)$
Math anxiety
$\displaystyle f'(x)= 6(8x^2+2)^5*(16x)=96x(8x^2+2)^5$
$\displaystyle g'(x)= 14(3x^2-9)^{13}*(6x)=84x(3x^2-9)^{13}$
$\displaystyle
[f(x)g(x)]' = f(x)g'(x)+g(x)f'(x)$
$\displaystyle =(8x^2+2)^6(84x)(3x^2-9)^{13}+(3x^2-9)^{14}(96x)(8x^2+2)^5
$
$\displaystyle =12x(8x^2+6)^5(3x^2-9)^{13}(7(8x^2+2)+8(3x^2-9))$
$\displaystyle =12x(8x^2+6)^5(3x^2-9)^{13}(56x^2+14+24x^2-72)$
$\displaystyle =12x(8x^2+6)^5(3x^2-9)^{13}(80x^2-58)$
$\displaystyle =24x(8x^2+6)^5(3x^2-9)^{13}(40x^2-29)$