How do you use the chain rule to differentiate $y=(3x^4-7x^3+3x^2-5x)^3$ ?

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Answer 1 · 2016-10-09T05:16:29

$(dy)/(dx)=3(12x^3-21x^2+6x-5)(3x^4-7x^3+3x^2-5x)^2$

Chain rule: $(dy)/(dx)=(dy)/(du)*(du)/(dx)$

Let $u=3x^4-7x^3+3x^2-5x$ ,
then $(du)/(dx)=12x^3-21x^2+6x-5$

$y=u^3$ , so $(dy)/(du)=3u^2=3(3x^4-7x^3+3x^2-5x)^2$

$:.(dy)/(dx)=(12x^3-21x^2+6x-5)*3(3x^4-7x^3+3x^2-5x)^2=3(12x^3-21x^2+6x-5)(3x^4-7x^3+3x^2-5x)^2$

How do you use the chain rule to differentiate y=(3x^4-7x^3+3x^2-5x)^3y=(3x^4-7x^3+3x^2-5x)^3?