Implicit Differentiation of $y^x=x^y$ Alright so my issue is that i get stuck at this point and do not know what i should do to isolate $\frac{dy}{dx}$ since it is asking for implicit differentiation
and this is what i have so far.$$\ln(y)*y^x*\frac{dy}{dx}=\ln(x)*x^y*\frac{dy}{dx}$$
 A: Try first taking the logarithm of both sides.
$$y^x=x^y$$
$$x \ln y=y \ln x$$
Next derive both sides with respect to $x$.
$$\frac{d}{dx}( x \ln y)=\frac{d}{dx}(y \ln x)$$
You'll need the product rule to continue. I'll do the LHS. Think you've got the RHS? Let me know!
$$\ln y + x \frac{1}{y} \frac{dy}{dx}=\frac{d}{dx}(y \ln x)$$

The answer ought to be equivalent to
$$\frac{dy}{dx} = \frac{y}{x}\cdot\frac{y-x \ln y}{x - y \ln x}$$
A: Take the logs and separate.
Because $\;y^x=x^y\;$, so then $\;\dfrac{\ln y}{y}=\dfrac{\ln x}{x}\;$ (unless $x=0\vee y=0$)
Now $\dfrac{\mathrm d\;}{\mathrm d x}\left(\dfrac{\ln x}{x}\right) = \dfrac{1-\ln x}{x^2}$
And you can take it from there.
A: By taking the natural log of both sides, we have $x \text{ln}(y) - y \text{ln} (x)=0$
Let $$x \text{ln}(y) - y \text{ln} (x)=0\equiv F$$.
By the Implicit Function Theorem (IFT), we obtain $$\frac{dy}{dx}= - \frac{\frac{\partial F }{\partial x}}{\frac{\partial F }{\partial y}}$$
$$ \implies \frac{dy}{dx} = \frac{\frac{y}{x} - \text{ln}(y)}{\frac{x}{y}-\text{ln}(x)} = \frac{y}{x}\cdot\frac{y-x \ln (y)}{x - y \ln(x)}.$$
