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anonymous
 one year ago
Find y′ if x^y = y^x.
anonymous
 one year ago
Find y′ if x^y = y^x.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well have you used implicit defrentiation before ? If you did you can take the natural logarithm for both sides

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ylnx = xlny \] You can apply implicit deffrentiation here, and then solve for y prime

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohhh true!! I wasn't thinking about logarithmic differentiation! Thanks so much!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry that's too late, it just took a while to upload

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Ahmadnedal That was very helpful!! I appreciate the help! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And I'm so happy to know that :)

idku
 one year ago
Best ResponseYou've already chosen the best response.0x^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
 one year ago
Best ResponseYou've already chosen the best response.0y`=[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)]
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