implicit differential problem

**fran1942** · December 3rd 2011, 02:30 PM

Hello, can someone please show me the steps involved in solving dy/dx for this function:
x + xy = 3.

My attempt is:
1 + y(dy/dx) + x = 0
=
dy/dx = (-1-x) / y

Thanks kindly.

**Quacky** · December 3rd 2011, 02:53 PM

Nearly.

$x+xy=3$

Using the product rule $\frac{d[uv]}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$ , we get:

$1+\underbrace{y\cdot\frac{d}{dx}[x]}+\underbrace{x\frac{d}{dx}[y]}=0$
I leave the $y$ , and differentiate the $x$ with respect to $x$ . I then leave the $x$ and differentiate the $y$ with respect to $x$ . I have to use the chain rule here:

$1+y+x\cdot\frac{dy}{dx}\cdot\frac{d}{dy}[y]=0$
Notice how the $dy$ 's would "cancel"? This is an application of the chain rule. I am now differentiating $y$ with respect to $y$ to give $1$ , leaving:

$1+y+x\frac{dy}{dx}=0$

Take it from here.

**Prove It** · December 3rd 2011, 03:57 PM

Originally Posted by fran1942

Hello, can someone please show me the steps involved in solving dy/dx for this function:
x + xy = 3.

My attempt is:
1 + y(dy/dx) + x = 0
=
dy/dx = (-1-x) / y

Thanks kindly.

Why should you have to use implicit differentiation here? Surely explicit differentiation would be easier...

$\displaystyle \begin{align*} x + x\,y &= 3 \\ x\,y &= 3 - x \\ y &= \frac{3 - x}{x} \\ y &= \frac{3}{x} - 1 \\ y &= 3x^{-1} - 1 \\ \frac{dy}{dx} &= -3x^{-2} \\ \frac{dy}{dx} &= -\frac{3}{x^2} \end{align*}$

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