How to solve this is implicit differentiation dy/dx? $x sin (y) + cos (x y) = x^{2}$ $xsin(y) + cos (xy) = x^2$

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How to solve this is implicit differentiation dy/dx? $x sin (y) + cos (x y) = x^{2}$ $xsin(y) + cos (xy) = x^2$

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Answer 1 · 2018-06-27T01:47:10

$\frac{d y}{d x} = \frac{2 x - sin (y) + y sin (x y)}{x cos (y) - x sin (x y)}$ $dy/dx = (2x-sin(y)+ysin(xy))/(xcos(y) - xsin(xy))$

Explanation:

Given: $x sin (y) + cos (x y) = x^{2}$ $xsin(y) + cos (xy) = x^2$

Differentiate all of the terms with respect to x:

$\frac{d (x sin (y))}{d x} + \frac{d (cos (x y))}{d x} = \frac{d (x^{2})}{d x} [1]$ $(d(xsin(y)))/dx + (d(cos (xy)))/dx = (d(x^2))/dx" [1]"$

For the first term, we must use the product rule,

$\frac{d (g h)}{d x} = \frac{d g}{d x} (h) + (g) \frac{d h}{d x}$ $(d(gh))/dx = (dg)/dx(h)+(g)(dh)/dx$ ,

where $g = x$ $g = x$ , $h = sin (y)$ $h=sin(y)$ , $\frac{d g}{d x} = 1$ $(dg)/dx = 1$ , and $\frac{d h}{d x} = cos (y) \frac{d y}{d x}$ $(dh)/dx = cos(y)dy/dx$ :

$\frac{d (x sin (y))}{d x} = (1) sin (y) + (x) cos (y) \frac{d y}{d x}$ $(d(xsin(y)))/dx = (1)sin(y)+(x)cos(y)dy/dx$

Substitute the above into equation [1]:

$sin (y) + (x) cos (y) \frac{d y}{d x} + \frac{d (cos (x y))}{d x} = \frac{d (x^{2})}{d x} [1.1]$ $sin(y)+(x)cos(y)dy/dx + (d(cos (xy)))/dx = (d(x^2))/dx" [1.1]"$

For the second term, we must use the chain rule:

$\frac{d (cos (x y))}{d x} = - sin (x y) \frac{d (x y)}{d x}$ $(d(cos (xy)))/dx = -sin(xy)(d(xy))/dx$

Then the product rule:

$\frac{d (cos (x y))}{d x} = - sin (x y) (y + x \frac{d y}{d x})$ $(d(cos (xy)))/dx = -sin(xy)(y+xdy/dx)$

Substitute the above equation into equation [1.1]:

$sin (y) + (x) cos (y) \frac{d y}{d x} - sin (x y) (y + x \frac{d y}{d x}) = \frac{d (x^{2})}{d x} [1.1]$ $sin(y)+(x)cos(y)dy/dx - sin(xy)(y+xdy/dx) = (d(x^2))/dx" [1.1]"$

Use the power rule for the last term:

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

Move all of the terms that do not contain $\frac{d y}{d x}$ $dy/dx$ to the right side:

$x cos (y) \frac{d y}{d x} - x sin (x y) \frac{d y}{d x} = 2 x - sin (y) + y sin (x y)$ $xcos(y)dy/dx - xsin(xy)dy/dx = 2x-sin(y)+ysin(xy)$

Factor out $\frac{d y}{d x}$ $dy/dx$ from the left side:

$(x cos (y) - x sin (x y)) \frac{d y}{d x} = 2 x - sin (y) + y sin (x y)$ $(xcos(y) - xsin(xy))dy/dx = 2x-sin(y)+ysin(xy)$

Divide both sides by the leading coefficient:

$\frac{d y}{d x} = \frac{2 x - sin (y) + y sin (x y)}{x cos (y) - x sin (x y)}$ $dy/dx = (2x-sin(y)+ysin(xy))/(xcos(y) - xsin(xy))$

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How to solve this is implicit differentiation dy/dx? $x sin (y) + cos (x y) = x^{2}$ $xsin(y) + cos (xy) = x^2$

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Explanation:

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How to solve this is implicit differentiation dy/dx? xsin(y)+cos(xy)=x2xsin(y) + cos (xy) = x^2

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How to solve this is implicit differentiation dy/dx? $x sin (y) + cos (x y) = x^{2}$ $xsin(y) + cos (xy) = x^2$