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Thread: find dy/dx

  1. #1
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    find dy/dx

    • find $\displaystyle \frac{dy}{dx}$ if $\displaystyle ln(xy)=x+y$


    this is what l tried to do to solve it:
    1. $\displaystyle ln(xy)=x+y$
    2. $\displaystyle lnx+lny=x+y$
    3. $\displaystyle y=lnx+lny-x$
    4. $\displaystyle y'=\frac{1}{x}+\frac{1}{y}-1$
    5. $\displaystyle y'=\frac{y+x}{xy}-1$
    6. $\displaystyle y'=\frac{y+x-xy}{xy}$

    however, the answer is suppose to be $\displaystyle \frac{xy-y}{x-xy}$

    how do you solve this?
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  2. #2
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    Quote Originally Posted by algebra2 View Post
    3. $\displaystyle y=lnx+lny-x$
    4. $\displaystyle y'=\frac{1}{x}+\color{red}\frac{1}{y}\color{black}-1$
    No. Remember that you are differentiating with respect to $\displaystyle x,$ so you must use the chain rule: $\displaystyle \frac d{dx}[\ln y]=\frac1y\frac{dy}{dx}.$

    For this problem, just use implicit differentiation, as follows.

    $\displaystyle \ln xy=x+y$

    $\displaystyle \Rightarrow\frac d{dx}\left[\ln xy\right]=\frac d{dx}\left[x+y\right]$

    $\displaystyle \Rightarrow\frac d{dx}\left[\ln x+\ln y\right]=1+\frac{dy}{dx}$

    $\displaystyle \Rightarrow\frac1x+\frac{dy/dx}y=1+\frac{dy}{dx}$

    $\displaystyle \Rightarrow\frac{dy/dx}y-\frac{dy}{dx}=1-\frac1x$

    $\displaystyle \Rightarrow\frac{dy}{dx}\left(\frac1y-1\right)=\frac{x-1}x$

    $\displaystyle \Rightarrow\frac{dy}{dx}=\frac{x-1}x\cdot\frac y{1-y}$

    $\displaystyle \Rightarrow\frac{dy}{dx}=\frac{xy-y}{x-xy}$
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