How do you find $\frac{d y}{d x}$ $(dy)/(dx)$ given $- x^{3} y^{2} + 4 = 5 x^{2} + 3 y^{3}$ $-x^3y^2+4=5x^2+3y^3$ ?

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How do you find $\frac{d y}{d x}$ $(dy)/(dx)$ given $- x^{3} y^{2} + 4 = 5 x^{2} + 3 y^{3}$ $-x^3y^2+4=5x^2+3y^3$ ?

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Answer 1 · 2016-11-22T20:46:12

$\frac{d y}{d x} = - \frac{10 x + 3 x^{2} y^{2}}{2 y x^{3} + 9 y^{2}}$ $dy/dx=-(10x +3x^2y^2)/(2yx^3+ 9y^2)$

Explanation:

$- x^{3} y^{2} + 4 = 5 x^{2} + 3 y^{3}$ $-x^3y^2 + 4= 5x^2+3y^3$
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$\Rightarrow - 3 x^{2} y^{2} + 2 y \frac{d y}{d x} \times (- x^{3}) = 10 x + 9 y^{2} \times \frac{d y}{d x}$ $rArr-3x^2y^2+2ydy/dxxx(-x^3)=10x + 9y^2xxdy/dx$
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$\Rightarrow - 3 x^{2} y^{2} - 2 y x^{3} \frac{d y}{d x} = 10 x + 9 y^{2} \times \frac{d y}{d x}$ $rArr-3x^2y^2-2yx^3dy/dx=10x + 9y^2xxdy/dx$
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$\Rightarrow - 2 y x^{3} \frac{d y}{d x} - 9 y^{2} \times \frac{d y}{d x} = 10 x + 3 x^{2} y^{2}$ $rArr-2yx^3dy/dx- 9y^2xxdy/dx=10x +3x^2y^2$
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$\Rightarrow \frac{d y}{d x} (- 2 y x^{3} - 9 y^{2}) = 10 x + 3 x^{2} y^{2}$ $rArrdy/dx(-2yx^3- 9y^2)=10x +3x^2y^2$
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$\Rightarrow \frac{d y}{d x} = \frac{10 x + 3 x^{2} y^{2}}{- 2 y x^{3} - 9 y^{2}}$ $rArrdy/dx=(10x +3x^2y^2)/(-2yx^3- 9y^2)$
$" "$
$\Rightarrow \frac{d y}{d x} = - \frac{10 x + 3 x^{2} y^{2}}{2 y x^{3} + 9 y^{2}}$ $rArrdy/dx=-(10x +3x^2y^2)/(2yx^3+ 9y^2)$

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How do you find $\frac{d y}{d x}$ $(dy)/(dx)$ given $- x^{3} y^{2} + 4 = 5 x^{2} + 3 y^{3}$ $-x^3y^2+4=5x^2+3y^3$ ?

1 Answer

Explanation:

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How do you find (dy)/(dx)dydx(dy)/(dx) given -x^3y^2+4=5x^2+3y^3−x3y2+4=5x2+3y3-x^3y^2+4=5x^2+3y^3?

1 Answer

Explanation:

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How do you find $\frac{d y}{d x}$ $(dy)/(dx)$ given $- x^{3} y^{2} + 4 = 5 x^{2} + 3 y^{3}$ $-x^3y^2+4=5x^2+3y^3$ ?