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

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

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Answer 1 · 2016-10-09T00:00:31

$\frac{d y}{d x} = 2 (2 x + 5) (x^{2} + 5 x) + 6 (3 x^{2} - 5) {(x^{3} - 5 x)}^{2}$ $(dy)/(dx)=2(2x+5)(x^2+5x)+6(3x^2-5)(x^3-5x)^2$

Explanation:

Chain rule: $\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $(dy)/(dx)=(dy)/(du)*(du)/(dx)$

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

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

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

Now adding both together,
$\frac{d y}{d x} = 2 (2 x + 5) (x^{2} + 5 x) + 6 (3 x^{2} - 5) {(x^{3} - 5 x)}^{2}$ $(dy)/(dx)=2(2x+5)(x^2+5x)+6(3x^2-5)(x^3-5x)^2$

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

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Explanation:

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