How do you use the chain rule to differentiate $f(x)=(3x-9)^2(4x^3+2x^-9)^-9$ ?

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Answer 1 · 2017-10-21T23:57:35

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$f(x)=(3x-9)^2(4x^3+2x^-9)^-9$

$f'(x)=(3x-9)^2d/dy(4x^3+2x^-9)^-9+(4x^3+2x^-9)^-9d/dx(3x-9)^2$

$=(3x-9)^2(-9)(4x^3+2x^-9)^-10d/dy(4x^3+2x^-9)+(4x^3+2x^-9)^-9(2)(3x-9)d/dx(3x-9)$

$=(3x-9)^2(-9)(4x^3+2x^-9)^-10(12x^2-18x^-10)+(4x^3+2x^-9)^-9(2)(3x-9)(3)$

$=(-9)(3x-9)^2(4x^3+2x^-9)^-10(12x^2-18x^-10)+(6)(4x^3+2x^-9)^-9(3x-9)$

How do you use the chain rule to differentiate f(x)=(3x-9)^2(4x^3+2x^-9)^-9f(x)=(3x-9)^2(4x^3+2x^-9)^-9?