[problem: Derivative Of The Natural Logarithm From The Definition | 1] Let $f : (0, \infty) \to \mathbb{R}$ be defined by $f(x) = \ln x$. Using only the definition of the derivative and the identity $\ln(1 + u)/u \to 1$ as $u \to 0$, show that $f'(a) = 1/a$ for every $a > 0$. [/problem]

[solution] **Step 1: Form the difference quotient.** For $a > 0$ and $h \neq 0$ with $a + h > 0$: \begin{align*} \frac{f(a+h) - f(a)}{h} &= \frac{\ln(a+h) - \ln a}{h} = \frac{1}{h} \ln\!\left(\frac{a+h}{a}\right) = \frac{1}{h} \ln\!\left(1 + \frac{h}{a}\right). \end{align*} The second equality uses the logarithm identity $\ln(x/y) = \ln x - \ln y$. **Step 2: Substitute $u = h/a$.** Since $a > 0$ is fixed, $h \to 0$ if and only if $u = h/a \to 0$, and $h = au$. Substituting: \begin{align*} \frac{1}{h} \ln\!\left(1 + \frac{h}{a}\right) &= \frac{1}{au} \ln(1 + u) = \frac{1}{a} \cdot \frac{\ln(1+u)}{u}. \end{align*} **Step 3: Take the limit.** As $h \to 0$ (equivalently $u \to 0$), the factor $\ln(1+u)/u \to 1$ by hypothesis. Therefore: \begin{align*} f'(a) &= \lim_{h \to 0} \frac{\ln(a+h) - \ln a}{h} = \frac{1}{a} \cdot 1 = \frac{1}{a}. \end{align*} [/solution]

[problem: Differentiability Of x Sin(1/x) And x Squared Sin(1/x) | 2] Define $f, g : \mathbb{R} \to \mathbb{R}$ by \begin{align*} f(x) &= \begin{cases} x \sin(1/x) & \text{if } x \neq 0, \\ 0 & \text{if } x = 0, \end{cases} \qquad g(x) = \begin{cases} x^2 \sin(1/x) & \text{if } x \neq 0, \\ 0 & \text{if } x = 0. \end{cases} \end{align*} 1. Show that $f$ is continuous at $0$ but not differentiable at $0$. 2. Show that $g$ is differentiable at $0$ with $g'(0) = 0$. 3. Show that $g'$ exists everywhere on $\mathbb{R}$ but is not continuous at $0$, concluding that $g$ is differentiable but not $C^1$. [/problem]

[solution] **Step 1: $f$ is continuous but not differentiable at $0$.** Continuity: for $x \neq 0$, $|f(x)| = |x \sin(1/x)| \leq |x| \to 0$ as $x \to 0$, so $f(x) \to 0 = f(0)$ by the squeeze theorem. Non-differentiability: the difference quotient at $0$ is \begin{align*} \frac{f(h) - f(0)}{h} &= \frac{h \sin(1/h)}{h} = \sin(1/h). \end{align*} As $h \to 0$, $\sin(1/h)$ oscillates without converging. Explicitly, $h_n = 1/(2\pi n + \pi/2)$ gives $\sin(1/h_n) = 1$, while $h_n' = 1/(2\pi n - \pi/2)$ gives $\sin(1/h_n') = -1$. Since the difference quotient takes values $1$ and $-1$ along sequences both converging to $0$, the limit does not exist. **Step 2: $g$ is differentiable at $0$ with $g'(0) = 0$.** The difference quotient at $0$ is: \begin{align*} \frac{g(h) - g(0)}{h} &= \frac{h^2 \sin(1/h)}{h} = h \sin(1/h). \end{align*} By the argument in Step 1 (continuity of $f$), $|h \sin(1/h)| \leq |h| \to 0$. Hence $g'(0) = 0$. **Step 3: $g'$ exists everywhere but is discontinuous at $0$.** For $x \neq 0$, the product and chain rules give: \begin{align*} g'(x) &= 2x \sin(1/x) + x^2 \cdot \cos(1/x) \cdot (-x^{-2}) = 2x \sin(1/x) - \cos(1/x). \end{align*} The term $2x \sin(1/x) \to 0$ as $x \to 0$, but $\cos(1/x)$ oscillates between $-1$ and $+1$: taking $x_n = 1/(2\pi n)$ gives $\cos(1/x_n) = 1$, so $g'(x_n) \to 0 - 1 = -1$, while taking $x_n' = 1/((2n+1)\pi)$ gives $\cos(1/x_n') = -1$, so $g'(x_n') \to 0 + 1 = 1$. Since $g'(x)$ does not converge as $x \to 0$, $g'$ is discontinuous at $0$ (recall $g'(0) = 0$). Thus $g$ is differentiable on all of $\mathbb{R}$, but $g' : \mathbb{R} \to \mathbb{R}$ is discontinuous at $0$. This proves that differentiability does not imply $C^1$: the class of differentiable functions is strictly larger than $C^1(\mathbb{R})$. [/solution]

[problem: Darboux's Theorem And A Derivative That Is Not Monotone | 3] 1. **(Darboux's theorem.)** Let $f : [a, b] \to \mathbb{R}$ be differentiable. Prove that $f'$ has the intermediate value property: if $f'(a) < \gamma < f'(b)$, then there exists $c \in (a, b)$ with $f'(c) = \gamma$. *Hint:* Apply the extreme value theorem to $g(x) = f(x) - \gamma x$. 2. Define $f(x) = x + 2x^2 \sin(1/x)$ for $x \neq 0$ and $f(0) = 0$. Show that $f'(0) = 1$ yet $f'$ takes negative values arbitrarily close to $0$. 3. Conclude that $f'(0) > 0$ does *not* imply that $f$ is increasing on any neighbourhood of $0$. [/problem]

[solution] **Step 1: Proof of Darboux's theorem.** Assume $f'(a) < \gamma < f'(b)$ (the reverse case is symmetric). Define $g(x) = f(x) - \gamma x$, which is continuous on $[a, b]$ and differentiable on $(a, b)$ with $g'(x) = f'(x) - \gamma$. By the extreme value theorem, $g$ attains its minimum at some $c \in [a, b]$. The minimum cannot occur at $a$: since $g'(a) = f'(a) - \gamma < 0$, the function $g$ is decreasing at $a$, so $g(a+h) < g(a)$ for small $h > 0$, meaning $a$ is not a minimiser. Similarly, the minimum cannot occur at $b$: since $g'(b) = f'(b) - \gamma > 0$, the function $g$ is increasing at $b$, so $g(b-h) < g(b)$ for small $h > 0$. Therefore $c \in (a, b)$. At an interior minimum of a differentiable function, the derivative vanishes (Fermat's theorem): $g'(c) = 0$, giving $f'(c) = \gamma$. **Step 2: $f'(0) = 1$ but $f' < 0$ near $0$.** At $x = 0$, the difference quotient gives: \begin{align*} \frac{f(h) - f(0)}{h} &= 1 + 2h \sin(1/h). \end{align*} Since $|2h \sin(1/h)| \leq 2|h| \to 0$, we get $f'(0) = 1$. For $x \neq 0$, the product and chain rules give: \begin{align*} f'(x) &= 1 + 4x \sin(1/x) - 2\cos(1/x). \end{align*} At $x_n = 1/(2\pi n)$: $\sin(1/x_n) = 0$ and $\cos(1/x_n) = 1$, so $f'(x_n) = 1 + 0 - 2 = -1$. Since $x_n \to 0$, the derivative $f'$ takes the value $-1$ arbitrarily close to $0$, despite $f'(0) = 1$. **Step 3: Positive derivative at a point does not imply local monotonicity.** If $f$ were increasing on some interval $(-\delta, \delta)$, the mean value theorem would give $f'(\eta) \geq 0$ for all points $\eta$ arising as MVT witnesses in that interval. But for any $\delta > 0$, choosing $n$ large enough gives $x_n = 1/(2\pi n) \in (0, \delta)$ with $f'(x_n) = -1 < 0$. Hence $f$ is not monotone increasing on any neighbourhood of $0$. The underlying issue is that $f'(0) > 0$ controls only the infinitesimal behaviour (the sign of $f(h) - f(0)$ for small $h$), not the behaviour of $f'$ at nearby points. Monotonicity on an interval requires $f' \geq 0$ on the *entire* interval, which is a much stronger condition than $f'(a) > 0$ at a single point. [/solution]

[problem: Taylor's Theorem And The Second Derivative Test | 2] Let $f : (a, b) \to \mathbb{R}$ be twice differentiable and let $c \in (a, b)$ with $f'(c) = 0$ and $f''(c) > 0$. 1. Use the definition of the derivative applied to $f'$ to show that there exists $\delta > 0$ such that $f'(x) < 0$ for $x \in (c - \delta, c)$ and $f'(x) > 0$ for $x \in (c, c + \delta)$. 2. Conclude that $f$ has a strict local minimum at $c$. 3. Show by example that the converse fails: exhibit a function with a strict local minimum at $c$, $f'(c) = 0$, but $f''(c) = 0$. [/problem]

[solution] **Step 1: Sign of $f'$ near $c$.** Since $f'(c) = 0$ and $f''(c) = \lim_{x \to c} (f'(x) - f'(c))/(x - c) = \lim_{x \to c} f'(x)/(x - c) > 0$, there exists $\delta > 0$ such that for all $x$ with $0 < |x - c| < \delta$: \begin{align*} \frac{f'(x)}{x - c} &> 0. \end{align*} For $x \in (c - \delta, c)$: $x - c < 0$, so $f'(x)/(x-c) > 0$ forces $f'(x) < 0$. For $x \in (c, c + \delta)$: $x - c > 0$, so $f'(x)/(x-c) > 0$ forces $f'(x) > 0$. **Step 2: $f$ has a strict local minimum at $c$.** From Step 1, $f$ is strictly decreasing on $(c - \delta, c)$ and strictly increasing on $(c, c + \delta)$. Precisely: for $x \in (c - \delta, c)$, the mean value theorem on $[x, c]$ gives $f(c) - f(x) = f'(\eta)(c - x)$ for some $\eta \in (x, c) \subseteq (c - \delta, c)$. Since $f'(\eta) < 0$ and $c - x > 0$, we get $f(c) - f(x) < 0$, i.e., $f(c) < f(x)$. Similarly, for $x \in (c, c + \delta)$, the MVT on $[c, x]$ gives $f(x) - f(c) = f'(\eta)(x - c) > 0$ (since $f'(\eta) > 0$ and $x - c > 0$), so $f(x) > f(c)$. Combining: $f(x) > f(c)$ for all $x \in (c - \delta, c + \delta) \setminus \{c\}$, so $c$ is a strict local minimum. **Step 3: Counterexample to the converse.** The function $f(x) = x^4$ has $f'(0) = 0$ and $f''(0) = 0$, yet $f$ has a strict global minimum at $0$ (since $x^4 > 0$ for $x \neq 0$). The second derivative test is inconclusive when $f''(c) = 0$: the function could have a minimum ($x^4$), a maximum ($-x^4$), or an inflection point ($x^3$), depending on the higher-order behaviour. [/solution]