Question

How do you find `(dy)/(dx)` given `x^2+y^2=1`?

Answer 1

`dy/dx=-x/y`

Explanation:

`x^2+y^2=1`

Differentiate wrt `x`:
`d/dxx^2+d/dxy^2=d/dx1`

We already know how to deal with the first and third terms, so lets get them out the way:
`d/dxx^2+d/dxy^2=d/dx1`
`:.2x+d/dxy^2=0`

For the remaining term we use the chain rule, we don't know how to differentiate `y^2` wrt `x` but we do know how to differentiate `y^2` wrt `y` (it the same as differentiating `x^2` wrt `x`!). The chain rule tells us that:
`dy/dx=dy/(du)*(du)/dx`, so we can rewrite:

`2x+d/dxy^2=0` as `2x+d/dy(y^2)*dy/dx=0`

We can know perform that final differentiation, as we are now differentiating a function of `y` wrt `y` so
`2x+d/dy(y^2)*dy/dx=0`
`:. 2x+2y*dy/dx=0`
We can then rearrange to get `dy/dx` as follows:
`2x+2y*dy/dx=0`
`:. 2y*dy/dx=-2x`

`:. dy/dx=-(2x)/(2y)`

`:. dy/dx=-x/y`