How do you implicitly differentiate $y = x y^{2} + x^{2} e^{x y}$ $y= xy^2 + x ^2 e^(x y)$ ?

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How do you implicitly differentiate $y = x y^{2} + x^{2} e^{x y}$ $y= xy^2 + x ^2 e^(x y)$ ?

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Answer 1 · 2016-03-01T05:26:13

$\frac{d y}{d x} = \frac{y^{2} + x^{2} y e^{x y} + 2 x e^{x y}}{1 - x^{3} e^{x y} - 2 x y}$ $frac{"d"y}{"d"x} = frac{y^2 + x^2ye^(xy) + 2xe^(xy)}{1 - x^3e^(xy) - 2xy}$

Explanation:

This problem requires knowledge of the Product Rule and the Chain Rule.

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

Differentiate both sides w.r.t. $x$ $x$ .

$\frac{d}{d x} (y) = \frac{d}{d x} (x y^{2} + x^{2} e^{x y})$ $frac{"d"}{"d"x}(y) = frac{"d"}{"d"x}(xy^2 + x^2e^(xy))$

$\frac{d y}{d x} = \frac{d}{d x} (x y^{2}) + \frac{d}{d x} (x^{2} e^{x y})$ $frac{"d"y}{"d"x} = frac{"d"}{"d"x}(xy^2) + frac{"d"}{"d"x}(x^2e^(xy))$

$\frac{d y}{d x} = (x \frac{d}{d x} (y^{2}) + y^{2} \frac{d}{d x} (x))$ $frac{"d"y}{"d"x} = (xfrac{"d"}{"d"x}(y^2) + y^2frac{"d"}{"d"x}(x))$

$+ (x^{2} \frac{d}{d x} (e^{x y}) + e^{x y} \frac{d}{d x} (x^{2}))$ $+ (x^2frac{"d"}{"d"x}(e^(xy)) + e^(xy)frac{"d"}{"d"x}(x^2))$

$\frac{d y}{d x} = (x \frac{d}{d y} (y^{2}) \frac{d y}{d x} + y^{2})$ $frac{"d"y}{"d"x} = (xfrac{"d"}{"d"y}(y^2)frac{"d"y}{"d"x} + y^2)$

$+ (x^{2} e^{x y} \frac{d}{d x} (x y) + e^{x y} (2 x))$ $+ (x^2e^(xy)frac{"d"}{"d"x}(xy) + e^(xy)(2x))$

$\frac{d y}{d x} = (x (2 y) \frac{d y}{d x} + y^{2})$ $frac{"d"y}{"d"x} = (x(2y)frac{"d"y}{"d"x} + y^2)$

$+ (x^{2} e^{x y} (y \frac{d}{d x} (x) + x \frac{d}{d x} (y)) + 2 x e^{x y})$ $+ (x^2e^(xy)(yfrac{"d"}{"d"x}(x) + xfrac{"d"}{"d"x}(y)) + 2xe^(xy))$

$\frac{d y}{d x} = (2 x y \frac{d y}{d x} + y^{2})$ $frac{"d"y}{"d"x} = (2xyfrac{"d"y}{"d"x} + y^2)$

$+ (x^{2} e^{x y} (y + x \frac{d y}{d x}) + 2 x e^{x y})$ $+ (x^2e^(xy)(y + xfrac{"d"y}{"d"x}) + 2xe^(xy))$

$\frac{d y}{d x} = 2 x y \frac{d y}{d x} + y^{2} + x^{2} y e^{x y} + x^{3} e^{x y} \frac{d y}{d x} + 2 x e^{x y}$ $frac{"d"y}{"d"x} = 2xyfrac{"d"y}{"d"x} + y^2 + x^2ye^(xy) + x^3e^(xy)frac{"d"y}{"d"x} + 2xe^(xy)$

Now make $\frac{d y}{d x}$ $frac{"d"y}{"d"x}$ the subject of formula.

$\frac{d y}{d x} - x^{3} e^{x y} \frac{d y}{d x} - 2 x y \frac{d y}{d x} = y^{2} + x^{2} y e^{x y} + 2 x e^{x y}$ $frac{"d"y}{"d"x} - x^3e^(xy)frac{"d"y}{"d"x} - 2xyfrac{"d"y}{"d"x} = y^2 + x^2ye^(xy) + 2xe^(xy)$

$\frac{d y}{d x} (1 - x^{3} e^{x y} - 2 x y) = y^{2} + x^{2} y e^{x y} + 2 x e^{x y}$ $frac{"d"y}{"d"x}(1 - x^3e^(xy) - 2xy) = y^2 + x^2ye^(xy) + 2xe^(xy)$

$\frac{d y}{d x} = \frac{y^{2} + x^{2} y e^{x y} + 2 x e^{x y}}{1 - x^{3} e^{x y} - 2 x y}$ $frac{"d"y}{"d"x} = frac{y^2 + x^2ye^(xy) + 2xe^(xy)}{1 - x^3e^(xy) - 2xy}$

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How do you implicitly differentiate $y = x y^{2} + x^{2} e^{x y}$ $y= xy^2 + x ^2 e^(x y)$ ?

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How do you implicitly differentiate $y = x y^{2} + x^{2} e^{x y}$ $y= xy^2 + x ^2 e^(x y)$ ?