If 4(x-y^3)-xy=16 ,evaluate dy/dx at the point (2,3)?

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If 4(x-y^3)-xy=16 ,evaluate dy/dx at the point (2,3)?

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Answer 1 · 2015-08-17T12:58:07

$\frac{d y}{d x} ∣_{x = 2, y = 3} = \frac{1}{110}$ $(dy)/dx|_(x=2,y=3) = 1/110$

Explanation:

$4 (x - y^{3}) - x y = 16$ $4(x-y^3)-xy = 16$
Differentiate with respect to $x$ $x$ :
$4 (1 - 3 y^{2} \frac{d y}{d x}) - y - x \frac{d y}{d x} = 0$ $4(1-3y^2(dy)/dx)-y-x(dy)/dx = 0$
$\frac{d y}{d x} = \frac{4 - y}{x + 12 y^{2}}$ $(dy)/dx = (4-y)/(x+12y^2)$

$\frac{d y}{d x} ∣_{x = 2, y = 3} = \frac{1}{110}$ $(dy)/dx|_(x=2,y=3) = 1/110$

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If 4(x-y^3)-xy=16 ,evaluate dy/dx at the point (2,3)?

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