How do you find the derivative of(x^5+x^6-8)^3?

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How do you find the derivative of(x^5+x^6-8)^3?

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Answer 1 · 2018-07-18T09:17:30

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Explanation:

$y = {(x^{5} + x^{6} - 8)}^{3}$ $y=(x^5+x^6-8)^3$

Let $u = x^{5} + x^{6} - 8$ $u=x^5+x^6-8$
then $\frac{d u}{d x} = 5 x^{4} + 6 x^{5}$ $(du)/(dx)=5x^4+6x^5$

Since $u = x^{5} + x^{6} - 8$ $u=x^5+x^6-8$ , then
$y = u^{3}$ $y=u^3$
$\frac{d y}{d u} = 3 u^{2}$ $(dy)/(du)=3u^2$

Therefore,
$\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ $(dy)/(dx)=(dy)/(du)times(du)/(dx)$
= $3 u^{2} \times (5 x^{4} + 6 x^{5})$ $3u^2times(5x^4+6x^5)$
= $3 {(x^{5} + x^{6} - 8)}^{2} (5 x^{4} + 6 x^{5})$ $3(x^5+x^6-8)^2(5x^4+6x^5)$

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How do you find the derivative of(x^5+x^6-8)^3?

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