# Did I solve this implicit differiation problem correctly?

If $x+2xy-y^2=2$, then at the point $\left(1,1\right)$, $\frac{\mathrm{d}y}{\mathrm{d}x} =$

differinate: $1 + 2y+2xy^{\prime} = 2y^{\prime}$

getting $y^{\prime}= \dfrac{1+2y}{2-2x}$

So when you put $\left(1,1\right)$ into the equation you get division by zero.

Thus DNE

My issue: $\left(1,1\right)$ is a solution to the primary equation, so this means that at the point in question the tagent line is horiziontal?

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Should your first equation after differentiation be: $1 + 2y + 2xy' = 0$? – James Evans Jan 19 '13 at 0:31
shouldn't $2xy$ result in $2y + 2xy^{\prime}$ – yiyi Jan 19 '13 at 0:34
You're correct, my mistake. – James Evans Jan 19 '13 at 0:36
The derivative of $y^2$ is $2yy'$, but since we are interested in $y=1$, by good luck this does not affect the conclusion. – André Nicolas Jan 19 '13 at 0:39
@AndréNicolas thanks for pointing out my mistake – yiyi Jan 19 '13 at 0:42

$$x + 2xy - y^2 = 2$$ Hence, $$\dfrac{d}{dx} \left(x + 2xy - y^2\right) = \dfrac{d(2)}{dx} = 0$$ Hence, $$\dfrac{dx}{dx} + \dfrac{d(2xy)}{dx} - \dfrac{d(y^2)}{dx} = 0$$ $$1 + 2y + 2x \dfrac{dy}{dx} - 2y\dfrac{dy}{dx} = 0 \implies \dfrac{dy}{dx} = \dfrac{1+2x}{2y-2x}$$ As you can see from your figure at $(1,1)$, the tangent is vertical and hence the slope is $\infty$.