Question

Find y′ if x^y = y^x.

Answer 1

Well have you used implicit defrentiation before ?
If you did you can take the natural logarithm for both sides

Answer 2

\[ylnx = xlny \]
You can apply implicit deffrentiation here, and then solve for y prime

Answer 3

Ohhh true!! I wasn't thinking about logarithmic differentiation! Thanks so much!

Answer 4

Welcome

Answer 5

Sorry that's too late, it just took a while to upload

Answer 6

@Ahmad-nedal That was very helpful!! I appreciate the help! :)

Answer 7

And I'm so happy to know that :)

Answer 8

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)]

Answer 9

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)]

Answer 10

Thanks @idku :)