Q1: If $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ are differentiable at $x_0$, is the function $f \circ g$ also differentiable at $x_0$?

A1: Yes!

A2: No, not in general.

Q2: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be differentiable at $x_0$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ be differentiable at $f(x_0)$. Which conclusion is correct?

A1: $f \circ g$ is differentiable at $x_0$

A2: $f \circ g$ is differentiable at $f(x_0)$

A3: $g \circ f$ is differentiable at $x_0$

A4: $g \circ f$ is differentiable at $f(x_0)$

A5: $g \circ f$ is differentiable at $g(x_0)$

Q3: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be differentiable at $x_0$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ be differentiable at $f(x_0)$. What is the correct formulation for the chain rule?

A1: $(f \circ g)^\prime(x_0) = f^\prime(x_0) \cdot g^\prime(x_0)$

A2: $(f \circ g)^\prime(x_0) = f^\prime(x_0) + g^\prime(x_0) + f(x_0)$

A3: $(g \circ f)^\prime(x_0) = f^\prime(x_0) \cdot g(x_0)$

A4: $(g \circ f)^\prime(x_0) = g^\prime( f(x_0) ) \cdot f^\prime(x_0)$

A5: $(g \circ f)^\prime(x_0) = g^\prime( f(x_0) ) \cdot g^\prime(x_0)$