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Math Help - How to derivate these?

  1. #1
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    How to derivate these?

    Hello, first year at university and a couple of odd looking functions are giving me a headache I have to derivate them so i can the value when x: 0.1,0.2,0.3, but my derivation so far has given me all the wrong answers and not the right ones

    Anyone bother to take a look at them please?

    How to derivate these?-math_image.aspx.gif


    and the other one:


    How to derivate these?-math_image.aspx.gif


    Thanks in advance
    Attached Thumbnails Attached Thumbnails How to derivate these?-math_image.aspx.gif  
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  2. #2
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    First, change the base \displaystyle \frac{e^{-x^2}}{\ln_{2}{x}} = \frac{e^{-x^2}\ln(2)}{\ln{x}}.

    Since \ln(2) is just a constant, you need to differentiate \displaystyle \frac{e^{-x^2}}{\ln{x}}.

    If \displaystyle y =  \frac{e^{-x^2}}{\ln{x}}, then:

    \ln{y} = \ln{e^{-x^2}-\ln\left(\ln{x}\right) = -x^2-\ln\left(\ln{x}\right).

    Differentiate both sides to get:

    \displaystyle \left(\frac{1}{y}\right)\frac{dy}{dx} = -2x-\frac{1}{x\ln{x}} = \frac{-\left(2x^2\ln{x}+1\right)}{x\ln{x}}

    \displaystyle \Rightarrow \frac{dy}{dx} = \frac{-\left(2x^2\ln{x}+1\right)}{x\ln{x}}y = \frac{-e^{-x^2}\left(2x^2\ln{x}+1\right)}{x\ln^2{x}}

    So your derivative is \displaystyle  \boxed{\frac{-\ln(2)e^{-x^2}\left(2x^2\ln{x}+1\right)}{x\ln^2{x}}} .

    Do the same for the second one!
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  3. #3
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    So your derivative is \displaystyle  \boxed{\frac{-\ln(2)e^{-x^2}\left(2x^2\ln{x}+1\right)}{x\ln^2{x}}} .
    Thanks for the fast reply, but i've never seen the ln^2, how is that imputted when i want to know the answer of 0.1? etc: 0.1 ln^2 0.1 , I just get a syn error on my calc ^^ any other way of typing it?

    managed to get the other one right by following your guidelines thanks
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  4. #4
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    Quote Originally Posted by Skolover View Post
    Thanks for the fast reply, but i've never seen the ln^2, how is that imputted when i want to know the answer of 0.1? etc: 0.1 ln^2 0.1 , I just get a syn error on my calc ^^ any other way of typing it?
    \ln^2{x} = (\ln{x})^2 = \ln(x)\ln(x), of course. So either enter \ln(0.1)\times \ln(0.1) or (\ln(0.1))^2 .
    managed to get the other one right by following your guidelines thanks
    No problem. I'm glad that I was of help.

    My 3th post.
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  5. #5
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    ah i see
    so \ln^2{x} = (\ln{(x}))^2
    are the same ^^ that'll come in handy :P new way to type it


    Congrats on your 300th post!
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  6. #6
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    I've just realised that 'do the same for the second one' was bit misleading, since taking
    logarithms makes the function more complicated to differentiate than it already were.

    We have y = \sin^{-1}({e^{2\ln{x}}})[/Math] [Math] = \sin^{-1}(x^2}). Using the chain rule gives:

    \displaystyle \frac{dy}{dx} = \frac{(x^2)'}{\sqrt{1-(x^2)^2}} = \frac{2x}{\sqrt{1-x^4}} =  \frac{2x}{\sqrt{1-x^4}}.

    Quote Originally Posted by Skolover View Post
    Congrats on your 300th post!
    Thanks.
    Last edited by TheCoffeeMachine; November 16th 2010 at 02:05 PM. Reason: Fixed a mistake!
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  7. #7
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    e^{2\ln{x}} = e^{\ln{x^2}} = x^2

    ... take the derivative of \sin^{-1}(x^2)
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