$$
(x^n)'= n x^{n-1}
$$

$$
(x^\infty)'=\infty x^{\infty-1}=\infty x^{\infty}
$$

$$
(x^n)'= n x^{n-1}
$$

$$
\left(\left(\frac xn\right)^n\right)'=\left(\frac{x}n\right)^{n-1}
$$

$$
\left(\left(\frac x\infty\right)^\infty\right)'=\left(\frac x\infty\right)^\infty
$$

if $$x=2$$, what is $\frac x\infty=0$ Then $0^\infty=0$

I can think about

$$
\left(\left(\frac {x+\infty}\infty\right)^\infty\right)'=\left(\frac {x+\infty}\infty\right)^\infty=e^x
$$

$$
\left(\frac {x+\infty}\infty\right)^\infty=\left(\frac {x}\infty+1\right)^\infty
$$

$$
\lim_{n\rightarrow \infty}\left(1+\frac xn\right)^n=e^x
$$

$$
e^x\approx x^\infty
$$

## Think about inverse function

What is the inverse function of $f(x)=x^n$?
$$
f^{-1}(x)=x^{\frac1n}
$$

Now if $f(x)=x^\infty$, what is the inverse function of $f(x)$

$$
f^{-1}(x)=x^0
$$

Then the derivative of $f^{-1}(x)$ should related to $x^{-1}$ in some sense.

If $f(x)=e^x$, it is like a polynomial of degree $\infty$

Then $f^{-1}(x)=\ln x$, it is like a polynomial of degree $0$

So the derivative of $\ln x$ should be something like $x^{-1}$.

## Let's try to do this step by step

$$
y=e^x=\left(\frac{x-\infty}\infty\right)^\infty
$$

$$
y^0=\frac{x-\infty}{\infty}
$$

$$
\infty y^0=x-\infty
$$

$$
x=\infty y^0+\infty
$$

Then the logarithm just looks like
$$\ln y= \infty y^0+\infty$$

$$\ln x= \infty x^0+\infty$$

When you take derivative
(reminder: $(x^n)'=nx^{n-1}$)
$$
(\ln x)'=\infty 0 y^{-1}=y^{-1}
$$
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