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anonymous

  • one year ago

Find y′ if x^y = y^x.

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  1. anonymous
    • one year ago
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    Well have you used implicit defrentiation before ? If you did you can take the natural logarithm for both sides

  2. anonymous
    • one year ago
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    \[ylnx = xlny \] You can apply implicit deffrentiation here, and then solve for y prime

  3. anonymous
    • one year ago
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    Ohhh true!! I wasn't thinking about logarithmic differentiation! Thanks so much!

  4. anonymous
    • one year ago
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    Welcome

  5. anonymous
    • one year ago
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  6. anonymous
    • one year ago
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    Sorry that's too late, it just took a while to upload

  7. anonymous
    • one year ago
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    @Ahmad-nedal That was very helpful!! I appreciate the help! :)

  8. anonymous
    • one year ago
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    And I'm so happy to know that :)

  9. idku
    • one year ago
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    x^y = y^x ln(x^y) = ln(y^x) yln(x)=xln(y) y` ln(x)+(y/x)=ln(y)+(y`/y) y` ln(x)-(y`/y)=ln(y)-(y/x) y`[ln(x)-(1/y)]=ln(y)-(y/x) y`=[ln(y)-(y/x)]/[ln(x)-(1/y)]

  10. idku
    • one year ago
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    y`=[ln(y)-(y/x)]/[ln(x)-(1/y)] (same line as the last in previous reply) y`=[ (xln(y)/x)-(y/x)]/[ln(x)-(1/y)] y`=[ (xln(y)-y)/x]/[ (yln(x)/y)-(1/y)] y`=[ (xln(y)-y)/x]/[ (yln(x)-1)/y] y`=[ y(xln(y)-y)]/[ x(yln(x)-1)]

  11. anonymous
    • one year ago
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    Thanks @idku :)

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