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Math Help - Differentiate with e^x

  1. #1
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    Differentiate with e^x

    Hi everyone need some help with my homework, I've been trying to solve these for awhile. Thanks

    1) y=xe^(-kx)


    2) f(x)= xe^(x)cscx
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  2. #2
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    this is my answer for number 1, but im a bit rusty, so dont take it as the correct answer.

    lny=lnx-kxlne

    lny=lnx -kx

    \frac{1}{y}\frac{dy}{dx}=\frac{1}{x}+-k-x\frac{dy}{dx}

    \frac{dy}{dx}(\frac{1}{y}+x)=\frac{1-kx}{x}

    \frac{dy}{dx}(\frac{1+xy}{y})=\frac{1-kx}{x}

    \frac{dy}{dx}=\frac{\frac{1-kx}{x}}{\frac{1+xy}{y}}

    \frac{dy}{dx}=\frac{y-kxy}{y+xy^2}

    \frac{dy}{dx}=\frac{(xe^{-kx})-kx(xe^{-kx})}{(xe^{-kx})+x(xe^{-kx})^2}

    \frac{dy}{dx}=\frac{1-kx}{1+x(xe^{-kx})}
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jpatrie View Post
    this is my answer for number 1, but im a bit rusty, so dont take it as the correct answer.

    lny=lnx-kxlne

    lny=lnx -kx

    {\color{red}\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}+-k-x\frac{dy}{dx}}

    \frac{dy}{dx}(\frac{1}{y}+x)=\frac{1-kx}{x}

    \frac{dy}{dx}(\frac{1+xy}{y})=\frac{1-kx}{x}

    \frac{dy}{dx}=\frac{\frac{1-kx}{x}}{\frac{1+xy}{y}}

    \frac{dy}{dx}=\frac{y-kxy}{y+xy^2}

    \frac{dy}{dx}=\frac{(xe^{-kx})-kx(xe^{-kx})}{(xe^{-kx})+x(xe^{-kx})^2}

    \frac{dy}{dx}=\frac{1-kx}{1+x(xe^{-kx})}
    this is overkill. logarithmic differentiation is not necessary and is too messy for this problem. simply using the product rule will suffice

    you also made a mistake where i put the red. so everything after that is wrong
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