Question

How do you implicitly differentiate `y^2/x= x^3 - 3yx^2`?

Answer 1

Use the product and quotients rules and do a lot of tedious algebra to get `dy/dx=(3x^4+2x^3y+y^2)/(2xy+x^4)`.

Explanation:

We will begin on the left hand side:
`y^2/x`

In order to take the derivative of this, we need to use the quotient rule:
`d/dx(u/v)=(u'v-uv')/v^2`

We have `u=y^2->u'=2ydy/dx` and `v=x->v'=1`, so:
`d/dx(y^2/x)=((2ydy/dx)(x)-(y^2)(1))/(x)^2`
`->d/dx(y^2/x)=(2xydy/dx-y^2)/x^2`

Now for the right hand side:
`x^3-3yx^2`

We can use the sum rule and multiplication of a constant rule to break this into:
`d/dx(x^3)-3d/dx(yx^2)`

The second of these will require the product rule:
`d/dx(uv)=u'v+uv'`
With `u=y->u'=dy/dx` and `v=x^2->v'=2x`. So:
`d/dx(x^3-3yx^2)=3x^2-((dy/dx)(x^2)+(y)(2x))`
`->d/dx(x^3-3yx^2)=3x^2-x^2dy/dx+2xy`

Our problem now reads:
`(2xydy/dx-y^2)/x^2=3x^2-x^2dy/dx+2xy`

We can add `x^2dy/dx` to both sides and factor out a `dy/dx` to isolate it:
`(2xydy/dx-y^2)/x^2=3x^2-x^2dy/dx+2xy`
`->(2xydy/dx)/x^2+x^2dy/dx-(y^2)/x^2=3x^2+2xy`
`->dy/dx((2xy)/x^2+x^2)=3x^2+2xy+(y^2)/x^2`
`->dy/dx=(3x^2+2xy+(y^2)/x^2)/((2xy)/x^2+x^2)`

I hope you like algebra, because this is one nasty equation that needs to be simplified:
`dy/dx=(3x^2+2xy+(y^2)/x^2)/((2xy)/x^2+x^2)`

`->dy/dx=((3x^4)/x^2+(2x^3y)/x^2+(y^2)/x^2)/((2xy)/x^2+x^4/x^2)`

`->dy/dx=((3x^4+2x^3y+y^2)/x^2)/((2xy+x^4)/x^2)`

`->dy/dx=(3x^4+2x^3y+y^2)/x^2*x^2/(2xy+x^4)`

`->dy/dx=(3x^4+2x^3y+y^2)/(2xy+x^4)`