Question #449c9

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Question #449c9

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Answer 1 · 2017-10-19T07:57:35

$\frac{d y}{d x} = \sqrt{1 - x^{3}} - \frac{3 x^{3}}{2 \sqrt{1 - x^{3}}}$ $dy/dx=sqrt(1-x^3)-(3x^3)/(2sqrt(1-x^3))$

Explanation:

We have $y = x \sqrt{1 - x^{3}}$ $y=xsqrt(1-x^3)$

We will differentiate both sides with respect to $x$ $x$ using the chain rule and the product rule.

The product rule is $\frac{d}{d x} (U \cdot V) = U \frac{d}{d x} V + V \frac{d}{d x} U$ $d/dx(U*V)=color(blue)Ud/dxcolor(red)V+color(blue)Vd/dxcolor(red)U$

$\frac{d}{d x} y = \frac{d}{d x} (x \sqrt{1 - x^{3}})$ $d/dxy=d/dx(xsqrt(1-x^3))$

$\frac{d y}{d x} = \sqrt{1 - x^{3}} \frac{d}{d x} x + x \frac{d}{d x} \sqrt{1 - x^{3}} \frac{d}{d x} 1 - x^{3}$ $dy/dx=color(blue)sqrt(1-x^3)d/dxcolor(red)x+color(blue)xd/dxcolor(red)sqrt(1-x^3)d/dxcolor(green)(1-x^3)$

$\frac{d y}{d x} = \sqrt{1 - x^{3}} \cdot 1 + x \frac{1}{2 \sqrt{1 - x^{3}}} \cdot (- 3 x^{2})$ $dy/dx=color(blue)sqrt(1-x^3)*color(red)1+color(blue)xcolor(red)(1/(2sqrt(1-x^3))*color(green)((-3x^2))$

$\frac{d y}{d x} = \sqrt{1 - x^{3}} - \frac{x 3 x^{2}}{2 \sqrt{1 - x^{3}}}$ $dy/dx=sqrt(1-x^3)-(x3x^2)/(2sqrt(1-x^3))$

$\frac{d y}{d x} = \sqrt{1 - x^{3}} - \frac{3 x^{3}}{2 \sqrt{1 - x^{3}}}$ $dy/dx=sqrt(1-x^3)-(3x^3)/(2sqrt(1-x^3))$

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Question #449c9

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