Question

Question #47376

Answer 1

`y'=-\frac{5x+3y}{3x+5y}`

Explanation:

`5x^2+5y^2+6xy=16`

`d/dx(5x^2+5y^2+6xy)=d/dx(16)`

The derivative of the right part is `0` because it's a constant. The derivative of the left part is:

`10x+10yy'+6y+6xy'`

So we have:
`10x+10yy'+6y+6xy'=0`

Let's reduce the equation:
`5x+5yy'+3y+3xy'=0`

Now let's factor `y'`
`5x+y'(5y+3x)+3y=0`

Solving for `y'`:
`y'(5y+3x)=-5x-3`

`y'=\frac{-5x-3y}{5y+3x}`

`y'=-\frac{5x+3y}{3x+5y}`