Q1: What is not a property of the exponential function $\exp: \mathbb{R} \rightarrow \mathbb{R}$?

A1: $\exp$ is continuous.

A2: $\exp(x+y) = \exp(x) \exp(y)$

A3: $\exp$ is strictly monotonically increasing.

A4: $\exp(2) = e^2$

A5: $\exp: \mathbb{R} \rightarrow [0,\infty)$ is bijective.

Q2: The logarithm function $\log: (0, \infty) \rightarrow \mathbb{R}$ is defined as the inverse function of $\exp: \mathbb{R} \rightarrow (0,\infty)$. Is $\log$ differentiable?

A1: No, nowhere!

A2: Yes, but only for $x_0 =1$.

A3: Yes, everywhere.

Q3: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be invertible and differentiable at $x_0$. What is the correct inversion formula if $f^{-1}$ is also differentiable at $y_0 = f(x_0)$?

A1: $ \displaystyle (f^{-1})^\prime(y_0) = f^\prime(x_0)$

A2: $ \displaystyle (f^{-1})^\prime(y_0) = \frac{1}{f^\prime(x_0)}$

A3: $ \displaystyle (f^{-1})^\prime(y_0) = \frac{1}{f^\prime(f^{-1}(x_0))}$

A4: $ \displaystyle (f^{-1})^\prime(y_0) = (f^\prime(f^{-1}(x_0)))^{-1}$

Q4: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be invertible and differentiable at $x_0$. What is the correct assumption to conclude that $f^{-1}$ is also differentiable at $y_0 = f(x_0)$?

A1: $f^{-1}$ is invertible at $y_0$

A2: $f^\prime(x_0) \neq 0$

A3: $f^\prime(x_0) \neq 0$ and $f^{-1}$ is continuous at $y_0$

A4: $f^\prime(x_0) \neq 0$ or $f^{-1}$ is continuous at $y_0$

A5: $f^\prime(x_0) = 0$ or $f^{-1}$ is continuous at $y_0$