anonymous
  • anonymous
Find y′ if x^y = y^x.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Well have you used implicit defrentiation before ? If you did you can take the natural logarithm for both sides
anonymous
  • anonymous
\[ylnx = xlny \] You can apply implicit deffrentiation here, and then solve for y prime
anonymous
  • anonymous
Ohhh true!! I wasn't thinking about logarithmic differentiation! Thanks so much!

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anonymous
  • anonymous
Welcome
anonymous
  • anonymous
anonymous
  • anonymous
Sorry that's too late, it just took a while to upload
anonymous
  • anonymous
@Ahmad-nedal That was very helpful!! I appreciate the help! :)
anonymous
  • anonymous
And I'm so happy to know that :)
idku
  • idku
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)]
idku
  • idku
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)]
anonymous
  • anonymous
Thanks @idku :)

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