Question

How do you differentiate `y^2cos(x) = x/(x-y)`?

Answer 1

See below.

Explanation:

In terms of differentiating the above equation, `dy/dx` is to be found.

If you differentiate both sides and fiddle a little bit, it's possible to extract `dy/dx`.

The left side will need to be differentiated with the product rule and the left with the quotient rule.

Furthermore, taking derivatives of functions of y occur as follows:

`(df(y))/dx=(df(y))/dy dy/dx`, by the chain rule.
This can be simply remembered as just differentiating the function with respect to `y` and just sticking a `dy/dx` on the end.

Let's differentiate the left side first:
`d/dx(y^2cosy)=d/dx(y^2)cosy+d/dx(cosy)y^2`
`d/dx(y^2cosy)=2ycosydy/dx-y^2sinydy/dx`

The right side:
`d/dx(x/(x-y))=(d/dx(x)(x-y)-d/dx(x-y)x)/(x-y)^2`
`d/dx(x/(x-y))=(x-y-x(1-dy/dx))/(x-y)^2`
`d/dx(x/(x-y))=(y+xdy/dx)/(x-y)^2`

Therefore:
`y^2cosy=x/(x-y)`
Becomes:
`2ycosydy/dx-y^2sinydy/dx=(y+xdy/dx)/(x-y)^2`
We need to group all of the `dy/dx` parts on one side in order to solve for it:
`2ycosydy/dx-y^2sinydy/dx-(xdy/dx)/(x-y)^2=y/(x-y)^2`
Factorise by `dy/dx`:
`dy/dx(2ycosy-y^2siny-x/(x-y)^2)=y/(x-y)^2`
Therefore (and this is a little messy):
`dy/dx=y/((x-y)^2(2ycosy-y^2siny-x/(x-y)^2))`
Tidy it up a little bit:
`dy/dx=y/((x-y)^2(2ycosy-y^2siny)-x)`