How do you use the chain rule to differentiate $y = \sqrt[5]{- x^{3} - 4}$ $y=root5(-x^3-4)$ ?

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Answer 1 · 2017-04-10T02:34:13

$\frac{d y}{d x} = \frac{- 3 x^{2}}{5 {(- x^{3} - 4)}^{- \frac{4}{5}}}$ $dy/dx=(-3x^2)/(5(-x^3-4)^(-4/5)$

Rewrite the function as,

$y = {(- x^{3} - 4)}^{\frac{1}{5}}$ $y=(-x^3-4)^(1/5)$

The chain rule has a general forumla of,

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

Applying that rule,

$\frac{d y}{d x} = ⎡ ⎢ ⎣ \frac{{(- x^{3} - 4)}^{- \frac{4}{5}}}{\frac{1}{5}} ⎤ ⎥ ⎦ \cdot (- 3 x^{2})$ $dy/dx=[(-x^3-4)^(-4/5)/(1/5)]*(-3x^2)$

Simplify,

$\frac{d y}{d x} = \frac{{(- x^{3} - 4)}^{- \frac{4}{5}} \cdot (- 3 x^{2})}{5}$ $dy/dx=[(-x^3-4)^(-4/5)*(-3x^2)]/5$

Rewrite (eliminate the negative exponents),

$\frac{d y}{d x} = \frac{- 3 x^{2}}{5 {(- x^{3} - 4)}^{- \frac{4}{5}}}$ $dy/dx=(-3x^2)/(5(-x^3-4)^(-4/5)$

How do you use the chain rule to differentiate y=5√−x3−4y=root5(-x^3-4)?