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

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Answer 1 · 2015-12-12T20:13:43

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

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.

How do you implicitly differentiate 1=-y^2/x+xy1=-y^2/x+xy?