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

Mar 1, 2016

$\frac{\text{d"y}{"d} x}{=} \frac{{y}^{2} + {x}^{2} y {e}^{x y} + 2 x {e}^{x y}}{1 - {x}^{3} {e}^{x y} - 2 x y}$

#### Explanation:

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

$y = x {y}^{2} + {x}^{2} {e}^{x y}$

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

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

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

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

$+ \left({x}^{2} \frac{\text{d"}{"d"x}(e^(xy)) + e^(xy)frac{"d"}{"d} x}{{x}^{2}}\right)$

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

$+ \left({x}^{2} {e}^{x y} \frac{\text{d"}{"d} x}{x y} + {e}^{x y} \left(2 x\right)\right)$

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

$+ \left({x}^{2} {e}^{x y} \left(y \frac{\text{d"}{"d"x}(x) + xfrac{"d"}{"d} x}{y}\right) + 2 x {e}^{x y}\right)$

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

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

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

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

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

$\frac{\text{d"y}{"d} x}{1 - {x}^{3} {e}^{x y} - 2 x y} = {y}^{2} + {x}^{2} y {e}^{x y} + 2 x {e}^{x y}$

$\frac{\text{d"y}{"d} x}{=} \frac{{y}^{2} + {x}^{2} y {e}^{x y} + 2 x {e}^{x y}}{1 - {x}^{3} {e}^{x y} - 2 x y}$