Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The graph

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

my answer:

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?

share|improve this question
    
Should your first equation after differentiation be: $1 + 2y + 2xy' = 0$? –  dirk5959 Jan 19 '13 at 0:31
    
shouldn't $2xy$ result in $2y + 2xy^{\prime}$ –  yiyi Jan 19 '13 at 0:34
1  
You're correct, my mistake. –  dirk5959 Jan 19 '13 at 0:36
2  
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

1 Answer 1

up vote 1 down vote accepted

$$x + 2xy - y^2 = 2$$ 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 $\infty$.

The plot was done using GeoGebra on Ubuntu. enter image description here

share|improve this answer
    
which program did you use to make the graph? –  yiyi Jan 19 '13 at 0:42
    
@MaoYiyi I made it on GeoGebra. –  user17762 Jan 19 '13 at 0:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.