Find y′ if x^y = y^x.

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Find y′ if x^y = y^x.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Well have you used implicit defrentiation before ? If you did you can take the natural logarithm for both sides
\[ylnx = xlny \] You can apply implicit deffrentiation here, and then solve for y prime
Ohhh true!! I wasn't thinking about logarithmic differentiation! Thanks so much!

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

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