# Use implicit differentiation to find dy/dx

$xy+x=2$

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

Thank you!

-
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 $y$? – apnorton Dec 14 '13 at 14:29

Taking the implicit derivative, we have:

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

Simplifying:

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

-
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 ${dy\over dx}=-{(1+y)\over x}.$

-

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!

-

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.

-