How do you differentiate ${(3 x + 4)}^{2} {(4 x - 1)}^{3}$ $(3x+4)^2(4x-1)^3$ using the power chain rule?

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Answer 1 · 2018-04-26T13:31:10

${(3 x + 4)}^{2} \cdot 12 {(4 x - 1)}^{2} + {(4 x - 1)}^{3} \cdot (18 x + 24)$ $(3x+4)^2*12(4x-1)^2 + (4x-1)^3*(18x+24)$

${(3 x + 4)}^{2} {(4 x - 1)}^{3}$ $(3x+4)^2(4x-1)^3$

Let

$f (x) = {(3 x + 4)}^{2}$ $f(x)=(3x+4)^2$ and $g (x) = {(4 x - 1)}^{3}$ $g(x) = (4x-1)^3$

The chain rule formula says

$f(x)*g'(x) + g(x)*f'(x)$

$g'(x)=3(4x-1)^2 (4)$ and $f'(x)=2(3x+4)(3)$

Simplifying a little, we get

$g'(x)=12(4x-1)^2$ and $f'(x)=6(3x+4)=18x+24$

Plugging these back into the chain rule formula above

$(3x+4)^2*12(4x-1)^2 + (4x-1)^3*(18x+24)$

You can continue to simplify this result if you desire, but this is as good a stopping point as any in my opinion.

How do you differentiate (3x+4)^2(4x-1)^3(3x+4)2(4x−1)3(3x+4)^2(4x-1)^3 using the power chain rule?