Find dy/ddx for xy² - sin(x+2y)= 2x?

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Find dy/ddx for xy² - sin(x+2y)= 2x?

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Answer 1 · 2015-08-20T14:13:49

$\frac{d y}{d x} = \frac{2 - y^{2} + cos (x + 2 y)}{2 x y + 2}$ $dy/dx=(2-y^2+cos(x+2y))/(2xy +2)$

Explanation:

Starting from the equation

$x y^{2} - sin (x + 2 y) = 2 x$ $xy^2-sin(x+2y)=2x$

We differentiate both sides of the equation, using $\frac{d}{d x}$ $d/dx$

$\frac{d}{d x} (x y^{2} - sin (x + 2 y)) = \frac{d}{d x} (2 x)$ $d/dx(xy^2-sin(x+2y))=d/dx(2x)$
$\frac{d}{d x} (x y^{2}) - \frac{d}{d x} (sin (x + 2 y)) = \frac{d}{d x} (2 x)$ $d/dx(xy^2)-d/dx(sin(x+2y))=d/dx(2x)$
$x \frac{d}{d x} (y^{2}) + y^{2} (1) - cos (x + 2 y) \frac{d}{d x} (x + 2 y) = 2$ $x d/dx(y^2) + y^2(1)-cos(x+2y)d/dx(x+2y)=2$
$x 2 y \frac{d y}{d x} + y^{2} - cos (x + 2 y) \frac{d}{d x} (x) + \frac{d}{d x} (2 y) = 2$ $x 2ydy/dx + y^2-cos(x+2y)d/dx(x)+d/dx(2y)=2$
$x 2 y \frac{d y}{d x} + y^{2} - cos (x + 2 y) + 2 \frac{d y}{d x} = 2$ $x 2ydy/dx + y^2-cos(x+2y)+2dy/dx=2$
$2 x y \frac{d y}{d x} + 2 \frac{d y}{d x} = 2 - y^{2} + cos (x + 2 y)$ $2xydy/dx +2dy/dx=2-y^2+cos(x+2y)$
$\frac{d y}{d x} (2 x y + 2) = 2 - y^{2} + cos (x + 2 y)$ $dy/dx(2xy +2)=2-y^2+cos(x+2y)$
$\frac{d y}{d x} = \frac{2 - y^{2} + cos (x + 2 y)}{2 x y + 2}$ $dy/dx=(2-y^2+cos(x+2y))/(2xy +2)$

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Find dy/ddx for xy² - sin(x+2y)= 2x?

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