find ${dy}/{dx}$ if $x^y + y^x = 1$ Find ${dy}/{dx}$ if $x^y + y^x = 1$.
I have no idea how to approach this problem. Can somebody please explain this to me?
 A: This may seem tricky but recall the derivatives of $2^x$ and $x^2$.
$$\frac{d}{dx}(2^x)=2^x\ln(2) \quad \frac{d}{dx}(x^2)=2x$$
Now, let's try approaching the problem.
$$\frac{d}{dx}(x^y)=yx^{y-1}+x^y\ln(x)\cdot\frac{dy}{dx},\qquad \frac{d}{dx}(y^x)=y^x\ln(y)+xy^{x-1}\cdot\frac{dy}{dx}$$
This may be hard to see but let's look at it another "way" (more of a physicist point of view):
Mulitply both sides by $dx$
$$d(x^y)=yx^{y-1}dx+x^y\ln(x)dy, \qquad d(y^x)=y^x\ln(y)dx + xy^{x-1}dy$$
As you can see, the derivative is of $x^y$ and $y^x$ is the derivative with respect to $x$ on the left side $+$ the derivative with repsect to $y$ on the right side.
So, 
$$\frac{dy}{dx}\left(x^y\ln(x)-xy^{x-1}\right)=-(y^x\ln(y)+yx^{y-1})$$

$$\frac{dy}{dx}=-\frac{y^x\ln(y)+yx^{y-1}}{x^y\ln(x)-xy^{x-1}}$$

A: There is no point $(x,y)$ in the first quadrant that satisfies $x^y+y^x=1$. Therefore the equation $x^y+y^x=1$ does not define implicitly a function $x\mapsto y=f(x)$, even locally.
A: $${d\over dx}x^y=yx^{y-1}+x^y\ln(x){dy\over dx}$$$${d\over dx}y^x=y^x\ln(y)+xy^{x-1}{dy\over dx}$$
Now solve.
A: Neither the real function nor accordingly its derivative can be defined. When the function itself cannot be defined its derivative by implicit differentiation is meaningless.
A: let denote $$F\left( x,y \right) =x^{ y }+y^{ x }-1$$ 
since  $$dF\left( x,y \right) =0\\ { F }_{ x }^{ \prime  }dx+{ F }_{ y }^{ \prime  }dy=0\\ { F }_{ x }^{ \prime  }+{ F }_{ y }^{ \prime  }\frac { dy }{ dx } =0\\ \frac { dy }{ dx } =-\frac { { F }_{ x }^{ \prime  } }{ { F }_{ y }^{ \prime  } } $$
so

$$\frac { dy }{ dx } =-\frac { { F }_{ x }^{ \prime  } }{ { F }_{ y }^{ \prime  } } =-\frac { y{ x }^{ y-1 }+{ y }^{ x }\ln { y }  }{ x{ y }^{ x-1 }+{ x }^{ y }\ln { x }  } $$

A: To find the derivative of function in the form $y=x^{f(x)}$, you need to know that $(\ln f(x))'= \frac{f'(x)}{f(x)}$, so $f'(x)=f(x)(\ln f(x))'$
Then differentiate both sides and think that $y = f(x)$ and $x$ is the independent variable, i.e. $$(x^y)'+(y^x)'=0.$$
$$(x^y)'=x^y(\ln x^y)'=x^y(y\ln x)'=x^y(y'\ln x+\frac{y}{x});\\
(y^x)'=y^x(\ln y^x)'=y^x(x\ln y)'=y^x(\ln y+x\frac{y'}{y}).
$$
So we need to pull out $y'$ from that:
$$x^y y'\ln x+x^y\frac{y}{x}+y^x\ln y+xy^x\frac{y'}{y}=0,\\
y'(x^y \ln x+\frac{xy^x}{y})=-x^y\frac{y}{x}-y^x\ln y.$$
Finally,
$$y'=-\frac{x^y\frac{y}{x}+y^x\ln y}{\frac{xy^x}{y}+x^y \ln x}.$$
