Question

How do you differentiate `y=x^2y-y^2-xy`?

Answer 1

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

Explanation:

Implicit differentiation is used when a function, here `y`, is not explicitly written in terms `x`. One will need the formula for differentiation of product & ratios of functions as well as chain formula to solve the given function.

For this, we take differential of both sides of the function `y=x^2y-y^2-xy` and differentiating we get

`(dy)/(dx)=d/(dx)(x^2y)-2y(dy)/(dx)-d/(dx)(xy)`

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

Transposing terms containing `(dy)/(dx)` to LHS we get

`(dy)/(dx)-x^2(dy)/(dx)+2y(dy)/(dx)+x(dy)/(dx)=2xy-y` or

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

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

Answer 2

Simplify then differentiate to find:

`(dy)/(dx) = 2x-1` when `y != 0`

Multiply by `y/(x^2-x-1)` to cover the case `y=0` and find:

`(dy)/(dx) = ((2x-1)y)/(x^2-x-1)`

Explanation:

graph{y=x^2y-y^2-xy [-10, 10, -5, 5]}

Notice that all of the terms are divisible by `y`, so do that first to find:

`1 = x^2-y-x`

with exclusion `y != 0`

We can rearrange this as:

`y = x^2-x-1`

Hence:

`(dy)/(dx) = 2x-1`

with exclusion `y=0`

What happens in the case `y=0`?

The original equation is satisfied, so its graph consists of the parabola `y = x^2-x-1` together with the x-axis `y=0` for which the derivative is `0`.

So to cover the case `y=0` we can multiply `2x-1` by `y/(x^2-x-1)`, since this has value `1` for all points on the curve where `y != 0` and value `0` when `y=0` (and `x^2-x-1 != 0`).

So: `(dy)/(dx) = ((2x-1)y)/(x^2-x-1)`