Question

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

Answer 1

`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 = xy^2 + x^2e^(xy)`

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

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

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

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

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

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

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

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

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

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

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

`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}` 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)`

`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^2ye^(xy) + 2xe^(xy)}{1 - x^3e^(xy) - 2xy}`