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Math Help - Check These Answers Please :)

  1. #1
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    Check These Answers Please :)

    <br />
y = x^{x^3}<br />
    <br />
y' = x^{2x^3} (1 - 3 ln x)<br />

    <br />
y = (cos 2x)^{\sqrt{x}}<br />
    <br />
y' = (cos 2x)^{\sqrt{x}} \left ( \frac{cos2xlncos2x - 4xsin2x}{2cos2x \sqrt{x}} \right)<br />
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  2. #2
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    Hello,
    Quote Originally Posted by qzno View Post
    <br />
y = x^{x^3}<br />
    <br />
y' = x^{2x^3} (1 - 3 ln x)<br />
    There are some problems.
    First of all, a^b a^c=a^{b+c} and not a^{bc}

    y=e^{x^3 \ln(x)}
    So :
    y'=(x^3 \ln(x))' e^{x^3 \ln(x)}=x^{x^3} (x^3 \ln(x))'

    Use product rule to calculate (x^3 \ln(x))' :
    =3x^2 \ln(x)+x^3 \cdot \frac 1x=3x^2 \ln(x)+x^2=x^2(3 \ln(x)+1)

    So y'=x^{x^3} \cdot x^2(3 \ln(x)+1)=x^{\color{red}x^3+2} (3 \ln(x)+1)

    Quote Originally Posted by qzno View Post

    <br />
y = (cos 2x)^{\sqrt{x}}<br />
    <br />
y' = (\cos (2x))^{\sqrt{x}} ( \frac{\cos(2x) \ln(\cos(2x)) - 4x \sin(2x)}{2\cos(2x) \sqrt{x}})<br />
    I changed some parts of the code

    Anyway, the last one looks perfect to me. Congratulations
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  3. #3
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    hahah thanks! do you mind checking some more? if not:

    <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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