Question

How do you implicitly differentiate `(x^2)/9+(x-y^2)/6=1`?

Answer 1

I found: `(dy)/(dx)=1/y((4x+3)/6)`

Explanation:

In this implicit expression we do not isolate `y` but we leave it as a function of `x`; when we differentiate we also need to differentiate `y` as well with respect to `x`. So, for example, if you have `y^2` differentiating you get: `2y(dy)/(dx)`.
In our case we get:
`1/9*2x+1/6(1-2y(dy)/(dx))=0`
`2/9x+1/6-1/3y(dy)/(dx)=0`
collect `(dy)/(dx)`:
`(dy)/(dx)=3/y(2/9x+1/6)`.

`(dy)/(dx)=1/y(2/3x+1/2)=1/y((4x+3)/6)`.