Finding dy/dx using implicit differentiation

Question

I'm having trouble getting the correct answer and any help would be appreciated.

Given a function: $2xy + y^2 = x + y$

Here are my steps:

1) Find the derivative of each value

$$2 + 2\frac{dy}{dx} + 2y\frac{dy}{dx} - 1 - 1\frac{dy}{dx} = 0$$

2) move $\frac{dy}{dx}$ to one side

$$1 = \frac{dy}{dx} (-2 -2y + 1)$$

3) isolate $\frac{dy}{dx} $

$$\frac{1}{-2 -2y + 1} = \frac{dy}{dx} $$

But the correct answer is: $$\frac{1 - 2y}{2x + 2y - 1}$$

Now obviously with simplifying the answer will differ a little, but I cannot seem where I went wrong.

used the product rule twice to find $2xy$'s derivative and the rest is simple exponential rules.

Thoughts?

Michael Hardy · Accepted Answer · 2014-10-05 22:20:13Z

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Use the product rule: $$ \frac{d}{dx}(2xy) = 2x\frac{dy}{dx} + 2\cdot1\cdot y. $$

answered Oct 5 '14 at 22:20

Michael Hardy

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mesel · Accepted Answer · 2014-10-05 22:20:36Z

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When you take derivative you will get

$$2xy'+2y+2yy'=1+y'$$

Be careful about $y^2\to2yy'$ and $2xy\to2y+2xy'$

answered Oct 5 '14 at 22:20

mesel

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$\begingroup$ So you leave the y variable in when taking the derivatives? ie: y = y * (dy/dx) ? $\endgroup$ – user1594121 Oct 5 '14 at 22:24
$\begingroup$ since $y$ depends on $x$, $f(y)'=f'(y).y'$ that is why $y^2\to 2yy'$ $\endgroup$ – mesel Oct 5 '14 at 22:25
$\begingroup$ ah the issue was 2xy -> 2y + 2xy'. My result was just 2y'. $\endgroup$ – user1594121 Oct 5 '14 at 22:28
$\begingroup$ $$(2xy)'=(2x)'y+2x(y)'=2y+2xy'$$ $\endgroup$ – mesel Oct 5 '14 at 22:30

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2 Answers 2