How do you find $\frac{d y}{d x}$ $(dy)/(dx)$ given $- 2 y^{2} + 3 = x^{3}$ $-2y^2+3=x^3$ ?

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How do you find $\frac{d y}{d x}$ $(dy)/(dx)$ given $- 2 y^{2} + 3 = x^{3}$ $-2y^2+3=x^3$ ?

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Answer 1 · 2018-06-10T16:42:50

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

Explanation:

Differentiate both sides as such:
$\frac{d}{d x} (- 2 y^{2} + 3) = \frac{d}{d x} (x^{3})$ $\frac{d}{dx}(-2y^{2}+3)=\frac{d}{dx}(x^{3})$
this leads to:

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

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

rearrange to make $\frac{d y}{d x}$ $\frac{dy}{dx}$ the factor to get:

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

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How do you find $\frac{d y}{d x}$ $(dy)/(dx)$ given $- 2 y^{2} + 3 = x^{3}$ $-2y^2+3=x^3$ ?

1 Answer

Explanation:

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