How do you implicitly differentiate y= xy^2 + x ^2 e^(x y) y=xy2+x2exy?

1 Answer
Mar 1, 2016

frac{"d"y}{"d"x} = frac{y^2 + x^2ye^(xy) + 2xe^(xy)}{1 - x^3e^(xy) - 2xy}dydx=y2+x2yexy+2xexy1x3exy2xy

Explanation:

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

y = xy^2 + x^2e^(xy)y=xy2+x2exy

Differentiate both sides w.r.t. xx.

frac{"d"}{"d"x}(y) = frac{"d"}{"d"x}(xy^2 + x^2e^(xy))ddx(y)=ddx(xy2+x2exy)

frac{"d"y}{"d"x} = frac{"d"}{"d"x}(xy^2) + frac{"d"}{"d"x}(x^2e^(xy))dydx=ddx(xy2)+ddx(x2exy)

frac{"d"y}{"d"x} = (xfrac{"d"}{"d"x}(y^2) + y^2frac{"d"}{"d"x}(x)) dydx=(xddx(y2)+y2ddx(x))

+ (x^2frac{"d"}{"d"x}(e^(xy)) + e^(xy)frac{"d"}{"d"x}(x^2))+(x2ddx(exy)+exyddx(x2))

frac{"d"y}{"d"x} = (xfrac{"d"}{"d"y}(y^2)frac{"d"y}{"d"x} + y^2) dydx=(xddy(y2)dydx+y2)

+ (x^2e^(xy)frac{"d"}{"d"x}(xy) + e^(xy)(2x))+(x2exyddx(xy)+exy(2x))

frac{"d"y}{"d"x} = (x(2y)frac{"d"y}{"d"x} + y^2) dydx=(x(2y)dydx+y2)

+ (x^2e^(xy)(yfrac{"d"}{"d"x}(x) + xfrac{"d"}{"d"x}(y)) + 2xe^(xy))+(x2exy(yddx(x)+xddx(y))+2xexy)

frac{"d"y}{"d"x} = (2xyfrac{"d"y}{"d"x} + y^2) dydx=(2xydydx+y2)

+ (x^2e^(xy)(y + xfrac{"d"y}{"d"x}) + 2xe^(xy))+(x2exy(y+xdydx)+2xexy)

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)dydx=2xydydx+y2+x2yexy+x3exydydx+2xexy

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

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)dydxx3exydydx2xydydx=y2+x2yexy+2xexy

frac{"d"y}{"d"x}(1 - x^3e^(xy) - 2xy) = y^2 + x^2ye^(xy) + 2xe^(xy)dydx(1x3exy2xy)=y2+x2yexy+2xexy

frac{"d"y}{"d"x} = frac{y^2 + x^2ye^(xy) + 2xe^(xy)}{1 - x^3e^(xy) - 2xy}dydx=y2+x2yexy+2xexy1x3exy2xy