Value of differentiation at a given point. If $x^y\cdot y^x=16$ then $\dfrac{dy}{dx}$ at $(2,2)$ is ?. After calling equation as $f(x)$ and differentiating I get $yx^{y-1}\cdot y^x+x^y\cdot y^x\cdot\ln(y)$ after plugging in value I get $16(1+\ln(2))$ but I don't think it's a right answer. Thanks. Hope you guys help.Also I would like to know what this graph is called if it has a name.
 A: If you define the implicit function $$F=x^y\, y^x-16=0$$ Compute the derivatives $$F'_x=x^{y-1} y^{x+1}+x^y y^x \log (y)$$ $$F'_y=x^{y+1} y^{x-1}+x^y y^x \log (x)$$ and from the implicit function theorem, after some minor simplifications, $$\frac{dy}{dx}=-\frac {F'_x}{F'_y}=-\frac{y (x \log (y)+y)}{x (y \log (x)+x)}$$ So, at a point where $y=x$, a beautiful result.
Edit
Just for your curiosity, $y$ has an analytical expression. If we consider the equation $x^y\, y^x=a$, the solution is given by $$y=\frac x{\log(x) }W\left(a^{\frac{1}{x}}\frac{ \log (x)}{x}\right)$$ where $W(z)$ is Lambert function.
A: Take logs and differentiate giving $$y'=-\frac{\frac{y}{x}+\ln y}{\frac{x}{y}+\ln x}$$ So at the point $(2,2)$ we have $y'=-1$. We could also have got that without any calculation since the curve is obviously symmetrical about the line $y=x$.

A: $$x^y\cdot y^x=16\tag{1}$$
Taking the derivative of both sides of $(1)$ via the product rule and using the chain rule we have $$\frac{\mathrm{d}}{\mathrm{d}x}\left(x\log y\right)e^{x\log y}x^y+\left(\frac{\mathrm{d}}{\mathrm{d}x}x^y\right)y^x=0$$
Using the chain rule again:
$$x^y\left(\log y + y^{\prime}(x)\frac{x}{y}\right)y^x+x^yy^x\left(\frac{y}{x}+y^{\prime}(x)\log x \right)=0$$
After some rearrangement:
$$y^{\prime}(x)=-\frac{y\left(y+x\log y\right)}{x(x+y\log x)}$$
The graph looks like:

But don't know about the name, sorry.
