Question

How do you implicitly differentiate `1=-y^2/x+xy`?

Answer 1

`dy/dx = (2xy)/ (2y-x^2)`

Explanation:

First, we will rearrange the equation:

`1=y^2/x = xy`

`1 + y^2 = x^2y`

In the above part, we've just sended the x to the right side and multiplied it.

Now, differentiating w.r.to x

`0 + 2y dy/dx = x^2 dy/dx + 2xy`

On the right hand in the above equation, we've applied the product rule

`2y dy/dx - x^2 dy/dx = 2xy`

Taking `dy/dx` common:

`dy/dx [2y - x^2] = 2xy`

`dy/dx = (2xy)/ (2y-x^2)`

The above equation will be our final answer.