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

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Answer 1 · 2016-10-09T00:00:31

$(dy)/(dx)=2(2x+5)(x^2+5x)+6(3x^2-5)(x^3-5x)^2$

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

We do this twice to derive both $(x^2+5x)^2$ and $2(x^3-5x)^3$

$d/(dx)(x^2+5x)^2$ : Let $u=x^2+5x$ , then $(du)/(dx)=2x+5$
$(dy)/(du)=2(x^2+5x)$
So $(dy)/(dx)=2(2x+5)(x^2+5x)$

$d/(dx)2(x^3-5x)^3$ : Let $u=x^3-5x$ , then $(du)/(dx)=3x^2-5$
$(dy)/(du)=6(x^3-5x)^2$
So $(dy)/(dx)=6(3x^2-5)(x^3-5x)^2$

Now adding both together,
$(dy)/(dx)=2(2x+5)(x^2+5x)+6(3x^2-5)(x^3-5x)^2$

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