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.