# Calculus Implicit Differentiation

I'm learning implicit differentiation and I've hit a snag with the following equation. $$f(x, y) = x + xy + y = 2$$ $$Dx(x) + Dx(xy) + Dx(y) = Dx(2)$$ $$1 + xy' + y + y' = 0$$ $$xy' + y' = -1 - y$$ $$y'(x + 1) = 1 + y$$ $$y' = \dfrac{(1 + y)}{(x + 1)}$$ $$y'' = \dfrac{(x + 1)y' - (1 + y)}{(x + 1)^2}$$ $$y'' = \dfrac{(x + 1)\dfrac{(1 + y)}{(x + 1)} - (1 + y)}{(x + 1)^2}$$

Ok now what according to this y'' = 0 which is wrong.

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Your mistake seems to originate when moving from this: $$xy' + y' = -1 - y$$ ...to this, where you "lost the sign": $$y'(x + 1) = 1 + y$$

We need $$y'(x + 1) = -(1 + y)$$

Let's back up: \begin{align} 1 + xy' + y + y' & = 0 \\ \\ xy' + y' & = -1 - y\\ \\ & = -(1 + y)\end{align}

Then we factor out $y'$ on the left-hand side, giving us: $$y'(x + 1) = -(1 + y)$$ Fixing for that, then, we get: \begin{align} y' &= \dfrac{-(1 + y)}{(x + 1)} \\ \\ y'' & = \frac{(x + 1)(-y') - [-(1 + y)]}{(x + 1)^2} \\ \\ & = \frac{-(x + 1)\dfrac{-(1 + y)}{(x + 1)} + (1 + y)}{(x + 1)^2} \\ \\ & = \frac{(1 + y) + (1+y)}{(x+1)^2}\\ \\ & = \frac{2(y+1)}{(x+1)^2} \end{align}

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Doesn't that work? -y - 1 = 0, -y = 1, 0 = 1 + y. – James Jul 16 '13 at 23:31
But $-y - 1\neq 0$. We have: $$1 + xy' + y + y' = 0 \iff xy' + y' = -1 - y = -(1 + y)$$ – amWhy Jul 16 '13 at 23:33
That makes sense, thanks. – James Jul 16 '13 at 23:36
Your welcome, James. – amWhy Jul 16 '13 at 23:41
@amWhy: Lots of typing ... +1 – Amzoti Jul 17 '13 at 1:11