Can we differentiate $x^y =-1$ ? Why?

Question

Suppose $x^y =4$. I find $y'$ as follows:$$ y\ln x = \ln 4\\ y'\ln x + \frac{y}{x} = 0\\ y' = -\frac{y}{x\ln x} \cdot $$

I am wondering how to find $y'$ when $x^y =-1$. Could you give some general rules/theorems?

Answer 1

We have that $x^y=e^{y \ln x}>0$ !

Answer 2

Negative numbers power to irrational (or some rationals) exponent is not define in real numbers. Ex (-1)^(0.5) Expression like x^y make sense only if x>0, and in this case there are no points (x,y) | x^y = -1.

Answer 3

The relation $y\log x =\log 4$ reveals that $$y' = -\frac{y}{x\log x} = -\frac{\log 4 / \log x}{x\log x} = -\frac{\log 4}{x\log^2 x}$$ which is what to expect since $y(x) = \log 4 /\log x$. Youtuber 3Blue1Brown has a nice video covering the intuition behind this.