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Math Help - integrals and derivatives

  1. #1
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    integrals and derivatives

    Hello everyone,

    Could you please correct the following?

    d/dx lnx=1/x

    d/dx 1/x=lnx

    int. e^-x= -1e^-x

    int. e^-2x= -2e^-2x

    d/dx e^-2x= -1/2e^-2x

    int. e^2x=2e^2x

    Thank you very much
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Hello everyone,

    Could you please correct the following?

    d/dx lnx=1/x
    yes

    d/dx 1/x=lnx
    no

    didn't you basically reverse what you did in the first question here? that's not taking the derivative, that's taking the anti-derivative

    hint: think of 1/x as x^{-1}

    int. e^-x= -1e^-x
    yes, but you forgot + C

    the form in which you wrote the answer makes me think you got it by coincidence though

    int. e^-2x= -2e^-2x
    it seems it was coincidence

    no, you took the derivative here. you want the integral.

    \int e^{kx}~dx = \frac 1ke^{kx} + C for k \ne 0 a constant

    d/dx e^-2x= -1/2e^-2x
    no, here you took the integral, not the derivative

    int. e^2x=2e^2x
    nope. see my comment two problems up


    you have been doing problems like these for a while now. you should be able to do them easily. maybe you forgot all the rules over your summer break? it happens, don't worry about it. but fix it, and fix it quick. please review your derivative and integration rules
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  3. #3
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    Hello, chocolatelover!

    Could you please correct the following?

    \frac{d}{dx}(\ln x) \:=\:\frac{1}{x} . . . . Right!

    \frac{d}{dx}\left(\frac{1}{x}\right) \:=\:\ln x . . . . no
    \frac{d}{dx}\left(\frac{1}{x}\right) \;=\;\frac{d}{dx}\left(x^{-1}\right) \;=\;-x^2 \;=\;-\frac{1}{x^2}


    \int e^{-x}dx \:=\: -e^{-x} + C . . . . Yes!

    \int e^{-2x}dx \:=\: -2e^{-2x}+ C . . . . no
    Let u \:=\:-2x\quad\Rightarrow\quad du \:=\:-2\,dx\quad\Rightarrow\quad dx \:=\:-\frac{du}{2}

    Substitute: . \int e^u\left(-\frac{du}{2}\right) \;=\;-\frac{1}{2}\int e^u\,du \:=\:-\frac{1}{2}\,e^u + C \;=\;-\frac{1}{2}\,e^{-2x}+C



    \frac{d}{dx}\left(e^{-2x}\right) \:=\:-\frac{1}{2}\,e^{-2x} +C . . . . Correct!

    \int e^{2x}dx \:=\:2e^{2x} + C . . . . no
    Let u = 2x \quad\Rightarrow\quad du = 2\,dx \quad\Rightarrow\quad dx = \frac{du}{2}

    Substitute: . \int e^u\left(\frac{du}{2}\right) \;=\;\frac{1}{2}\int e^u\,du \;=\;\frac{1}{2}\,e^u + C \;=\;\frac{1}{2}\,e^{2x} + C

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  4. #4
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    Thank you very much

    Regards
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