Determine y' if y^x = x^y. (I took ln of both sides)

Question

jennychan12 · Answer

i did 
xlny = ylnx
1/y(x) + lny = y(1/x) + lnx
sorry. maybe i'm just being stupid. but what do i do next?

anonymous · Answer

let 

y^x=x^y

take natural log of both sides 

xln(y)=yln(x)

differentiate with product rule 
ln(y)+x/y *y'=y/x+ln(x)y'
just simplify it now .

anonymous · Answer

$$x\ln(y)=y\ln(x)$$
take derivative wrt x and get 
$$\ln(y)+\frac{xy'}{y}=y'\ln(x)+\frac{y}{x}$$

jennychan12 · Answer

oh wait. i think i just forgot to put dy/dx.
lol.

jennychan12 · Answer

wait. i did that but i think i'm getting the wrong answer.
yeah i got what you wrote @satellite73

jennychan12 · Answer

oh wait. ohhh so then you just move the dy/dx's over
so i got dy/dx = [y(xlny-y)]/[x(x-ylnx)]

jennychan12 · Answer

$$\frac{ dy }{ dx } = \frac{ y(x \ln y-y) }{ x( x- y  \ln x)  }$$

jennychan12 · Answer

so then just simplify right?

jennychan12 · Answer

ok i got the answer