1. ## implicit differential problem

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.

2. ## Re: implicit differential problem

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.

3. ## Re: implicit differential problem

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*}