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Math Help - implicit differential problem

  1. #1
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    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.
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  2. #2
    Super Member Quacky's Avatar
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    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.
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  3. #3
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    Re: implicit differential problem

    Quote Originally Posted by fran1942 View Post
    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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