Question

2x^2+y^2=8xy, find the dy/dx.

Answer 1

`(dy)/(dx)=(4y-2x)/(y-4x)`

Explanation:

`(d)/(dx)(2x^2)=4x`

`(d)/(dx)(y^2)=2y(dy)/(dx)`

`(d)/(dx)(8xy)`

`=8(d)/(dx)(xy)`

`=8((d)/(dx)(x)y+x(d)/(dx)(y))`

`=8[1y+x(dy)/(dx)]`

`=8y+8x(dy)/(dx)`

Let's put it altogether now:

`2x^2+y^2=8xy`

Differentiating both sides gives:

`4x+2y(dy)/(dx)=8y+8x(dy)/(dx)`

We are solving for dy/dx so we need to gather those terms on one side and the terms without on the opposing side:

I'm going to first subtract 4x on both sides:

`2y(dy)/(dx)=8y-4x+8x(dy)/(dx)`

I'm not going to subtract 8xdy/dx on both sides:

`2y(dy)/(dx)-8x(dy)/(dx)=8y-4x`

It is time to factor the dy/dx out of the two terms on the left:

`(dy)/(dx)(2y-8x)=8y-4x`

Divide both sides by (2y-8x):

`(dy)/(dx)=(8y-4x)/(2y-8x)`

Reduce right hand side fraction:

`(dy)/(dx)=(4y-2x)/(y-4x)`

Answer 2

`(8y-4x)/(2y-8x)`

Explanation:

Differentiate implicitly with respect to x

noting that

`(d)/(dx)` (y² ) = 2y
`(dy)/(dx)`

Differentiate 8xy using the product rule

Given

2x² + y² = 8xy, then

4x + 2y
`(dy)/(dx)` = 8x
`(dy)/(dx)` + 8y

Collect terms in
`(dy)/(dx)`

2y
`(dy)/(dx)` - 8x
`(dy)/(dx)` = 8y - 4x

`(dy)/(dx)` (2y - 8x) = 8y - 4x

`(dy)/(dx)` =
`(8y-4x)/(2y-8x)`