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.

$xy+x=2$

I know the answer is $-(1+y)\over x$, but I don't know how to solve to get the answer.

Thank you!

share|improve this question
    
Are you looking for an explicit solution for $\frac{dy}{dx}$ in terms of only $x$, or do you are if the solution contains a $y$? –  anorton Dec 14 '13 at 14:29

4 Answers 4

Taking the implicit derivative, we have:

$$y + x y' + 1 = 0$$

Simplifying:

$$y' = -\dfrac{1}{x} - \dfrac{y}{x} = -\dfrac{1+y}{x}$$

share|improve this answer
    
As simple as that! +1 –  amWhy Dec 15 '13 at 0:06

Given $xy+x=2$ we can differentiate implicitly with respect to $x$. So ${dy\over dx}(xy+x)={dy\over dx}(2)$. This gives us $x{dy\over dx}+y+1=0$. Solving for ${dy\over dx}$ we obtain ${dy\over dx}=-{1\over x}-{y\over x}$. Thus ${dy\over dx}=-{(1+y)\over x}.$

share|improve this answer

Implicit differentiation works just like regular differentiation--you take the derivative of everything with respect to $x$. However, when you take the derivative of $y$ for example, you don't just $\frac{d}{dx}\left(y\right)=1$ like normally because the rate at which $y$ changes could depend on $x$ (and often does)! So we need to take that into account, this is done by including $\frac{dy}{dx}$, the rate at which $y$ changes with respect to $x$! So for example, the derivative of $y^2$ with respect to $x$ is $\frac{d}{dx}\left(y^2\right)=2y \cdot \frac{dy}{dx}$.

Now back to the problem at hand. You wanted to be able to find the slopes at $(x,y)$ on the graph created by $xy+x=2$. Slopes mean derivative, but we have a $y$ in the equation, so we use implicit differentiation. $$ \frac{d}{dx}\left(xy\right)+\frac{d}{dx}\left(x\right)=\frac{d}{dx}\left(2\right) $$ which is $$ 1\cdot y+x\cdot\left(1 \cdot \frac{dy}{dx}\right)+1=0 $$ So we have $$ x\frac{dy}{dx}+y+1=0 $$ since the slopes, $\frac{dy}{dx}$ are what we want, we solve for that. Which yields: $$ \frac{dy}{dx}=\frac{-1-y}{x}=-\frac{y+1}{x} $$ Now notice given a point $(x,y)$ on the graph, we can plug them in to find the slope $\frac{dy}{dx}$ on the graph! Moreover, given a point and a slope, we can find the tangent line to the graph there!

share|improve this answer

Let $f(x,y)=xy+x-2=0$

An application of the chain rule now yields $$\frac {\partial f(x,y)}{\partial x}=x\frac{\partial y}{\partial x}+{(y+1)}=0$$ Solving for $\frac{\partial y}{\partial x}$ you can verify your answer.

share|improve this answer

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.