How do you use the chain rule to differentiate $y=(3x-1)(-3x^2-4)^-3$ ?

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Answer 1 · 2017-12-13T01:05:14

$dy/dx = 3*[15x^2 -6x + 4]/(-3x^2-4)^4$

$dy/dx = (d[(3x-1)(-3x^2-4)^-3])/dx$

$dy/dx = (3x-1)*(d[(-3x^2-4)^-3])/dx + ((-3x^2-4)^-3 ) (d(3x-1))/dx$

$dy/dx = 3*(-3x^2-4)^-3 + (3x-1)*(d[(-3x^2-4)^-3])/dx$

$dy/dx = 3*(-3x^2-4)^-3 + (3x-1)$ $[-3*((-3x^2-4)^-4)*(d(-3x^2-4))/dx]$

$dy/dx = 3*(-3x^2-4)^-3 +18x*(3x-1)*(-3x^2-4)^-4$

$dy/dx = 3/(-3x^2-4)^3 +18x*(3x-1)/(-3x^2-4)^4$

$dy/dx = [3*(-3x^2-4) +18x*(3x-1)]/(-3x^2-4)^4$

$dy/dx = [-9x^2 + 12 +54x^2-18x]/(-3x^2-4)^4$

$dy/dx = [45x^2 -18x + 12]/(-3x^2-4)^4$

$dy/dx = 3*[15x^2 -6x + 4]/(-3x^2-4)^4$

How do you use the chain rule to differentiate y=(3x-1)(-3x^2-4)^-3y=(3x-1)(-3x^2-4)^-3?