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Math Help - find dy/dx

  1. #1
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    find dy/dx

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


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

    however, the answer is suppose to be \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. y=lnx+lny-x
    4. y'=\frac{1}{x}+\color{red}\frac{1}{y}\color{black}-1
    No. Remember that you are differentiating with respect to x, so you must use the chain rule: \frac d{dx}[\ln y]=\frac1y\frac{dy}{dx}.

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

    \ln xy=x+y

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

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

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

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

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

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

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