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Math Help - check :)

  1. #1
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    Oct 2008
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    109

    check :)

    <br />
y = (1 + 4e^{2x})^{2x}<br />
    <br />
y' = (1 + 4e^{2x})^{2x} \left( \frac{16xe^{2x} + 8e^{2x}(1 + 4e^{2x})ln(1 + 4e^{2x})}{(1 + 4e^{2x})}\right)<br />

    <br />
y = \left( \frac{1}{x} \right)^{ln x}<br />
    <br />
y' = \left( \frac{1}{x} \right)^{ln x} \left(\frac{x^2lnx + ln \frac{1}{x}}{x}\right)<br />

    <br />
y = \frac {\sqrt[3]{x^3 + 1}}{x^3(x-1)^3}<br />
    <br />
y' = \frac {\sqrt[3]{x^3 + 1}}{x^3(x-1)^3} \left( \frac{x^2}{x^3+1} - \frac{3}{x} + \frac{3}{x-1} \right)<br />

    <br />
y = \frac {ln x}{e^{\sqrt {x}}cos^4x}<br />
    <br />
y' = \frac {ln x}{e^{\sqrt {x}}cos^4x} \left( \frac{1}{xlnx} - \frac{e^{\sqrt{x}}}{2 \sqrt{x} e^{\sqrt{x}}} - 4tanx \right)<br />
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  2. #2
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    anything?
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  3. #3
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    Hello, qzno!

    y \:=\: (1 + 4e^{2x})^{2x}

    y' \:= \:(1 + 4e^{2x})^{2x} \left( \frac{16xe^{2x} + 8e^{2x}(1 + 4e^{2x})\ln(1 + 4e^{2x})}{(1 + 4e^{2x})}\right)
    I have a "2" where you have "8".


    y \:= \:\left(\frac{1}{x}\right)^{\ln x}

    I did it differently and got an entirely different-looking answer.

    Watch this . . . y \;=\;\left(x^{-1}\right)^{\ln x} \;=\;(x)^{-\ln x}

    Take logs: . \ln(y) \;=\;\ln(x)^{-\ln x} \;=\;-\ln x\cdot\ln x \;=\;-(\ln x)^2

    Then: . \frac{1}{y}\cdot y' \;=\;-2\cdot\ln x \cdot \frac{1}{x} \;=\;\frac{-2\ln x}{x}

    Therefore: . y' \;=\;\left(\frac{1}{x}\right)^{\ln x}\left(\frac{-2\ln x}{x}\right)



    In the last two, be careful with your algebra.
    . . You missed a "minus" in both of them.


    y \:= \:\frac {\sqrt[3]{x^3 + 1}}{x^3(x-1)^3}
    We have: . y \;=\;\frac{(x^3+1)^{\frac{1}{3}}}{x^3(x-1)^3}

    Take logs: . \ln(y) \;=\;\ln\left[\frac{(x^3+1)^{\frac{1}{3}}}{x^3(x-1)^3}\right]

    . . =\;\ln(x^3+1)^{\frac{1}{3}} - \ln\left[x^3(x-1)^3\right]

    . . =\;\ln(x^3+1)^{\frac{1}{3}} - \underbrace{{\color{red}\bigg[}\ln(x^3) + \ln(x-1)^3{\color{red}\bigg]}}_{\text{minus }both\text{ of them!}}

    . . = \;\ln(x^3+1)^{\frac{1}{3}} - \ln(x^3)\: {\color{red}-} \ln(x-1)^3


    Got it?

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  4. #4
    Member
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    Oct 2008
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    109
    hahah thanks! ive been doing these for soo long today its not funny
    is that last one right?
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