How do you find $\frac{d y}{d x}$ $(dy)/(dx)$ given ${(2 x - 3)}^{2} + {(4 y - 5)}^{2} = 10$ $(2x-3)^2+(4y-5)^2=10$ ?

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How do you find $\frac{d y}{d x}$ $(dy)/(dx)$ given ${(2 x - 3)}^{2} + {(4 y - 5)}^{2} = 10$ $(2x-3)^2+(4y-5)^2=10$ ?

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Answer 1 · 2016-11-06T13:56:37

The derivative is $\frac{d y}{d x} = \frac{3 - 2 x}{8 y - 10}$ $dy/dx = (3 - 2x)/(8y - 10)$ .

Explanation:

Start by expanding the parentheses.

$4 x^{2} - 12 x + 9 + 16 y^{2} - 40 y + 25 = 10$ $4x^2 - 12x + 9 + 16y^2 - 40y + 25 = 10$

$\frac{d}{d x} (4 x^{2} - 12 x + 9 + 16 y^{2} - 40 y + 25) = \frac{d}{d x} (10)$ $d/dx(4x^2 - 12x + 9 + 16y^2 - 40y + 25) = d/dx(10)$

$\frac{d}{d x} (4 x^{2}) + \frac{d}{d x} (- 12 x) + \frac{d}{d x} (9) + \frac{d}{d x} (16 y^{2}) + \frac{d}{d x} (- 40 y) + \frac{d}{d x} (25) = \frac{d}{d x} (10)$ $d/dx(4x^2) + d/dx(-12x) + d/dx(9) + d/dx(16y^2) + d/dx(-40y) + d/dx(25) = d/dx(10)$

$8 x - 12 + 0 + 32 y (\frac{d y}{d x}) - 40 (\frac{d y}{d x}) + 0 = 0$ $8x - 12 + 0 + 32y(dy/dx) - 40(dy/dx) + 0 = 0$

$32 y (\frac{d y}{d x}) - 40 (\frac{d y}{d x}) = 12 - 8 x$ $32y(dy/dx) - 40(dy/dx) = 12 - 8x$

$\frac{d y}{d x} (32 y - 40) = 12 - 8 x$ $dy/dx(32y - 40) = 12 - 8x$

$\frac{d y}{d x} = \frac{12 - 8 x}{32 y - 40}$ $dy/dx = (12 - 8x)/(32y - 40)$

$\frac{d y}{d x} = \frac{4 (3 - 2 x)}{4 (8 y - 10)}$ $dy/dx = (4(3 - 2x))/(4(8y - 10))$

$\frac{d y}{d x} = \frac{3 - 2 x}{8 y - 10}$ $dy/dx= (3- 2x)/(8y - 10)$

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How do you find $\frac{d y}{d x}$ $(dy)/(dx)$ given ${(2 x - 3)}^{2} + {(4 y - 5)}^{2} = 10$ $(2x-3)^2+(4y-5)^2=10$ ?

1 Answer

Explanation:

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1 Answer

Explanation:

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How do you find $\frac{d y}{d x}$ $(dy)/(dx)$ given ${(2 x - 3)}^{2} + {(4 y - 5)}^{2} = 10$ $(2x-3)^2+(4y-5)^2=10$ ?