What is the implicit derivative of $1 = x y e^{y} - x cos (x y)$ $1= xye^y-xcos(xy)$ ?

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What is the implicit derivative of $1 = x y e^{y} - x cos (x y)$ $1= xye^y-xcos(xy)$ ?

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Answer 1 · 2016-07-16T10:22:52

$\frac{d y}{d x} = \frac{cos (x y) - x y sin (x y) - y e^{y}}{x e^{y} + x y e^{y} + x^{2} sin (x y)}$ $(dy)/(dx) = (cos(xy) - xysin(xy) - ye^y)/(xe^y+xye^y+x^2sin(xy))$

Explanation:

$\frac{d}{d x} (1) = \frac{d}{d x} (x y e^{y}) - \frac{d}{d x} (x cos (x y))$ $d/(dx)(1) = d/(dx)(xye^y) - d/(dx)(xcos(xy))$

To do these derivatives, remember $\frac{d}{d x} (y) = \frac{d y}{d x}$ $d/(dx)(y) = (dy)/(dx)$ and hammer your product and chain rules.

$0 = y e^{y} + x \frac{d y}{d x} e^{y} + x y \frac{d y}{d x} e^{y} - cos (x y) + x (y + x \frac{d y}{d x}) sin (x y)$ $0 = ye^y + x(dy)/(dx)e^y + xy(dy)/(dx)e^y - cos(xy) + x(y+x(dy)/(dx))sin(xy)$

$\frac{d y}{d x} (x e^{y} + x y e^{y} + x^{2} sin (x y)) = cos (x y) - x y sin (x y) - y e^{y}$ $(dy)/(dx)(xe^y+xye^y+x^2sin(xy)) = cos(xy) - xysin(xy) - ye^y$

$\frac{d y}{d x} = \frac{cos (x y) - x y sin (x y) - y e^{y}}{x e^{y} + x y e^{y} + x^{2} sin (x y)}$ $(dy)/(dx) = (cos(xy) - xysin(xy) - ye^y)/(xe^y+xye^y+x^2sin(xy))$

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What is the implicit derivative of $1 = x y e^{y} - x cos (x y)$ $1= xye^y-xcos(xy)$ ?

1 Answer

Explanation:

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Science

Math

Humanities

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What is the implicit derivative of 1= xye^y-xcos(xy) 1=xyey−xcos(xy)1= xye^y-xcos(xy) ?

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Explanation:

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What is the implicit derivative of $1 = x y e^{y} - x cos (x y)$ $1= xye^y-xcos(xy)$ ?