Question

How do you implicitly differentiate `2= e^(xy^2-xy)-y^2x^3+y`?

Answer 1

See the answer below:

Answer 2

`dy/dx = (3x^2y^2-e^(xy^2-xy)(y^2-y))/(e^(xy^2-xy)(2xy-x)-2x^3y+1)`

Explanation:

There is another (often faster) approach using partial derivatives. Suppose we cannot find `y` explicitly as a function of `x`, only implicitly through the equation `F(x, y) = 0` which defines `y` as a function of `x, y = y(x)`. Therefore we can write `F(x, y) = 0` as `F(x, y(x)) = 0`. Differentiating both sides of this, using the partial chain rule gives us

`(partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y))`

We have:

`e^(xy^2-xy)-x^3y^2+y=2`

So Let `F(x,y) = e^(xy^2-xy)-x^3y^2+y-2`; Then;

`(partial F)/(partial x) = e^(xy^2-xy)(y^2-y)-3x^2y^2`

`(partial F)/(partial y) = e^(xy^2-xy)(2xy-x)-2x^3y+1`

And so:

`dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y)`
`\ \ \ \ \ \ = - (e^(xy^2-xy)(y^2-y)-3x^2y^2)/(e^(xy^2-xy)(2xy-x)-2x^3y+1 )`
`\ \ \ \ \ \ = (3x^2y^2-e^(xy^2-xy)(y^2-y))/(e^(xy^2-xy)(2xy-x)-2x^3y+1)`