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  1. #1
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    derivative

    what is the derivative of -3/x^2 + 3lnx - 6e^x

    thank you for your help
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Derivative of

    :: bx^a = bax^(a-1)

    :: aln(x) = a/x

    :: ae^(x) = ae^(x)
    -------------------------

    Use them Derivative of

    f(x) + g(x) + k(x)
    is
    f'(x) + g'(x) + k'(x)
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  3. #3
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    so its -6e^x + 3/x.

    i have problems w the first one becasue its a fraction.

    now if i leave the fraction its -3/ 2x. but should i leave the fraction or not?

    or is the end result -3/ 2x - 6e^x + 3/x ?
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  4. #4
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by makaveli89 View Post
    what is the derivative of -3/x^2 + 3lnx - 6e^x

    thank you for your help
    It can be done as

    Derivative of

    -3/x^2 = -3x^(-2)

    is given by

    -3*-2 x^(-3)
    = 6 x^(-3)

    -----------------------

    Answer thus will be

    = 6 x^(-3) -6e^x + 3/x.
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  5. #5
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    Quote Originally Posted by makaveli89 View Post
    what is the derivative of -3/x^2 + 3lnx - 6e^x
    Your formatting is ambiguous. I think the other replies have assumed that you meant the following:

    . . . . . -\frac{3}{x^2}\, +\, 3\ln(x)\, -\, 6e^x

    Is the above what you meant? Or did you mean "-3/(x^2 + 3ln(x) - 6e^x)", which, in LaTeX, looks like the following?

    . . . . . \frac{-3}{x^2\, +\, 3\ln(x)\, -\, 6e^x}

    When you reply, please show what you have tried so far, so we can "see" which part is causing confusion. Thank you!
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