Implicit Differentiation Quesrion $$x^y = y^x$$ is the equation
My solution is :$$\frac { dy }{ dx } =\left( \ln { y-\frac { y }{ x }  }  \right) /\left( \ln { x-\frac { x }{ y }  }  \right) $$
Just wondering if this is correct!
 A: Both are correct. Notice that
$$\frac{\left( \ln { y-\frac { y }{ x }  }  \right)}{\left( \ln { x-\frac { x }{ y }  }  \right)} = \frac{x\ln y - y}{x} \frac{y}{y\ln x - x}$$ By the nature of the function we have $x \ln y = y \ln x$, therefore
$$\frac{dy}{dx} = \frac{y\ln x - y}{x}\frac{y}{x\ln x - x} = \frac{\ln x - 1}{x^2}\frac{y^2}{\ln y - 1}$$
A: Your solution does not match mine:
$$
x^y = y^x \implies x\ln y = y\ln x \implies \frac{\ln y}{y} = \frac{\ln x}{x} \\
\frac{\frac1y\cdot y-1\cdot\ln y }{y^2}dy = \frac{\frac1x\cdot x-1\cdot\ln x }{x^2}dx\\
\frac{1-\ln y}{y^2} dy= \frac{1-\ln x}{x^2}dx\\
\frac{dy}{dx} = \frac{1-\ln x}{1-\ln y}\cdot\frac{y^2}{x^2} 
$$
As I said, that does not seem to reduce to your answer.
A: Here is a much more intuitive and rigorous approach:
The logarithmic function is defined as the geometric area under a $\frac 1t$ vs. $t$ curve, mathematically equivalent to: $f(y) = log(y) = \int_{1}^{y} \frac 1t \ dt$, bounded by $[1,y]$ and the line $y=0$. (As a note, in mathematics $log(y)$ is equivalent to $ln(y)$, and thus $log(y)$ by default has base $e$.)
We can extend this notion further by simply taking the inverse of $log(y)$, which yields the exponential function notated as $exp(x)$ or its equivalent algebraic form, $e^x$. Furthermore, this is analogous to defining a function, let's say $g(x) = b^x$ for some base $b\in \mathbb{R}$, by setting $b^x = e^{xlogb}$. (Note that by taking the natural logarithm [log] of both sides of the equality, with respect to its argument x, immediately proves this claim).
Now that we have a basic understanding of base conversion, the above question can be evaluated as follows:
$x^y = y^x \Rightarrow e^{ylogx}=e^{xlogy}$
Via implicit differentiation, with respect to the argument x, we have:
$e^{ylogx}(\frac {dy}{dx}logx+\frac yx)=e^{xlogy}(logy+\frac xy\frac {dy}{dx}) \Rightarrow \frac{dy}{dx}(x^ylogx-\frac {y^xx}{y})=(y^xlogy-\frac {x^yy}{x})$
Via division and referring back to the initial condition $x^y=y^x$ we have:
$\frac {dy}{dx}=\frac {y^xlogy-\frac {x^yy}{x}}{x^ylogx-\frac{y^xx}{y}}=\frac {logy - \frac yx}{logx- \frac xy}$
Thus your solution is correct. 
