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Thread: 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.

    $\displaystyle x+xy=3$

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

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

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

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

    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 \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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