Chapter 4: Differentiation
11/19 problems solved
Exercise 4.1
Since \(\lim_{x \to c} f(x) = L\), we know that for all \(\varepsilon > 0\), there exists \(\delta > 0\) such that for all \(x \in U_\delta(c)\), \(L - \varepsilon < f(x) < L + \varepsilon\).
Multiplying by \(k > 0\), we get \(kL - k\varepsilon < k \cdot f(x) < kL + k\varepsilon\). For \(\varepsilon' = k\varepsilon > 0\), we know there exists \(\delta > 0\) where all \(x \in U_\delta(c)\) satisfies \(kL - \varepsilon' < k \cdot f(x) < kL + \varepsilon'\). Therefore, \(kf\) converges to \(kL\) as \(x\) approaches \(c\) for \(k > 0\). If \(k < 0\), then changing \(\varepsilon'\) to \(-k\varepsilon\) maintains the same inequality. When \(k = 0\), it is trivial to show the same inequality is true. Therefore, \(kf\) converges to \(kL\) as \(x\) approaches \(c\).
Exercise 4.2
We know that for all \(\varepsilon > 0\), there exists \(\delta > 0\) such that for any \(x \in U_\delta(c)\), \(-\varepsilon < |f(x)| < \varepsilon\). If \(f(x) > 0\), then \(-\varepsilon < f(x) < \varepsilon\). Otherwise, \(-\varepsilon < -f(x) < \varepsilon\) which means \(\varepsilon > f(x) > -\varepsilon\). Therefore, \(-\varepsilon < f(x) < \varepsilon\) so \(\lim_{x \to c} f(x) = 0\).
Exercise 4.3
If \(\lim_{x \to c} f(x) = L\), then for any \(\varepsilon > 0\) there exists \(\delta > 0\) such that for all \(x \in U_\delta(c)\), \(L - \varepsilon < f(x) < L + \varepsilon\). The same holds for \(\lim_{x \to 0} f(x + c) = L\) except for all \(x \in U_\delta(0)\), \(L - \varepsilon < f(x + c) < L + \varepsilon\). We can write any \(x \in U_\delta(c)\) as a \(x' \in U_\delta(0)\) plus \(c\). We can also go the other way for any \(x \in U_\delta(0)\) written as a \(x' \in U_\delta(c)\) minus \(c\). Therefore, the two claims are equivalent.
Exercise 4.4
Exercise 4.5
Exercise 4.6
(a)
\[ \begin{align} (f + g)'(c) &= \lim_{x \to c} \frac{(f + g)(x) - (f + g)(c)}{x - c} \\ &= \lim_{x \to c} \frac{f(x) + g(x) - f(c) - g(c)}{x - c} \\ &= \lim_{x \to c} \frac{f(x) - f(c)}{x - c} + \lim_{x \to c} \frac{g(x) - g(c)}{x - c} \\ &= f'(c) + g'(c) \\ \end{align} \]
(b)
\[ \begin{align} (kf)'(c) &= \lim_{x \to c} \frac{(kf)(x) - (kf)(c)}{x - c} \\ &= \lim_{x \to c} \frac{k \cdot f(x) - k \cdot f(c)}{x - c} \\ &= k \cdot \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \\ &= k f'(c) \\ \end{align} \]
Exercise 4.7
(a)
We can show that \((x^n)' = nx^{n-1}\) for \(n > 0\) using induction. When \(n = 1\), we need to prove that \((x)' = 1\). This is simple since \(\lim_{t \to x} \frac{t - x}{t - x} = 1\). Now we assume \((x^n)' = nx^{n-1}\) and must show that \((x^{n+1})' = (n+1)x^n\). We can do this by following the following equalities \((x^{n+1})' = (x \cdot x^n)' = (x)' \cdot x^n + x \cdot (x^n)' = x^n + x \cdot nx^{n-1} = (n+1)x^n\). Therefore, \((x^n)' = nx^{n-1}\) for \(n > 0\).
(b)
(c)
Exercise 4.8
\[ \begin{align} \frac{d}{dx} \sqrt{x + \sqrt{x + \sqrt{x}}} &= \frac{1}{2}\left(x + \sqrt{x + \sqrt{x}}\right)^{-1/2}\left(\frac{d}{dx}\left(x + \sqrt{x + \sqrt{x}}\right)\right) \\ &= \frac{1}{2}\left(x + \sqrt{x + \sqrt{x}}\right)^{-1/2}\left(1 + \frac{1}{2}(x + \sqrt{x})^{-1/2}\right)\left(\frac{d}{dx}\left(x + \sqrt{x}\right)\right) \\ &= \frac{1}{2}\left(x + \sqrt{x + \sqrt{x}}\right)^{-1/2}\left(1 + \frac{1}{2}(x + \sqrt{x})^{-1/2}\right)\left(1 + \frac{1}{2}x^{-1/2}\right) \\ \end{align} \]
Exercise 4.9
False, consider \(f(x) = |x|\) and \(g(x) = -|x|\). Both functions are not differentiable at \(x = 0\). However, \(f(x) + g(x) = |x| - |x| = 0\) is differentiable at \(x = 0\).
Exercise 4.10
If \(f(x) = g(x) + C\), then the derivative of \(f\) is \(g'(x) + 0 = g'(x)\). Therefore, \(f'(x) = g'(x)\).
If \(f'(x) = g'(x)\), then \(f'(x) - g'(x) = 0\). Since the derivative of \(f(x) - g(x)\) is \(0\), then \(f(x) - g(x) = C\) for some constant \(C\) since the field of real numbers is complete and satisfies the Zero Derivative Property. Therefore, \(f(x) = g(x) + C\).
Exercise 4.11
Exercise 4.12
Exercise 4.13
Exercise 4.14
Since \(x_1 < x_2\), there exists \(c\) such that \(x_2 = x_1 + c\). Moreover, since \(x \in [x_1, x_2]\), there exists \(x = x_1 + d\).
Suppose for the sake of contradiction that there exist \(\lambda, \lambda' \in [0, 1]\) such that \(x = \lambda x_1 + (1 - \lambda) x_2\) and \(x = \lambda' x_1 + (1 - \lambda') x_2\). Rewriting \(x\) and \(x_2\) above and simpifying gives us \(d = x_1 + (1 - \lambda) c\) and \(d = x_1 + (1 - \lambda') c\). These two equations show us that \(\lambda = \lambda'\).
Exercise 4.15
\[ \begin{align} f(x) &\leq \frac{x_2 - x}{x_2 - x_1} f(x_1) + \frac{x - x_1}{x_2 - x_1} f(x_2) \\ (x_2 - x_1) f(x) &\leq (x_2 - x) f(x_1) + (x - x_1) f(x_2) \\ (x_2 - x + x - x_1) f(x) &\leq (x_2 - x) f(x_1) + (x - x_1) f(x_2) \\ (x_2 - x) f(x) + (x - x_1) f(x) &\leq (x_2 - x) f(x_1) + (x - x_1) f(x_2) \\ (x_2 - x) f(x) - (x_2 - x) f(x_1) &\leq (x - x_1) f(x_2) - (x - x_1) f(x) \\ (x_2 - x) (f(x) - f(x_1)) &\leq (x - x_1) (f(x_2) - f(x)) \\ \frac{f(x) - f(x_1)}{x - x_1} &\leq \frac{f(x_2) - f(x)}{x_2 - x} \\ \end{align} \]
Exercise 4.16
First, we show that \(\min\{x_1, \dotsc, x_n\} \leq x\). Since \(x = \lambda_1 x_1 + \dotsb + \lambda_n x_n\), we can rewrite \(x\) so the inequality is \(\min\{x_1, \dotsc, x_n\} \leq \lambda_1 x_1 + \dotsb + \lambda_n x_n\). Let the minimum of \(x_1, \dotsc, x_n\) be \(c\). Since \(c \leq x_k\) for all \(1 \leq k \leq n\), we can write \(d_k = x_k - c > 0\). Substituting \(x_k\) in the original inequality shows us \(c \leq \lambda_1 (c + d_1) + \dotsb + \lambda_n (c + d_n)\) which means \(c \leq (\lambda_1 + \dotsb + \lambda_n) c + (\lambda_1 d_1 + \dotsb + \lambda_n d_n)\). Since \(\lambda_1 + \dotsb + \lambda_n = 1\), \(c \leq c + \lambda_1 d_1 + \dotsb + \lambda_n d_n\) and so this inequality is true.
The argument to show that \(x \leq \max\{x_1, \dotsc, x_n\}\) follows a similar structure. We let \(c\) be the maximum of \(x_1, \dotsc, x_n\) and since \(x_k \leq c\) for all \(1 \leq k \leq n\), we can write \(d_k = c - x_k > 0\). This gives us \(\lambda_1 (c - d_1) + \dotsb + \lambda_n (c - d_n) \leq c\) so \((\lambda_1 + \dotsb + \lambda_n) c - (\lambda_1 d_1 + \dotsb + \lambda_n d_n) \leq c\) and so \(c - (\lambda_1 d_1 + \dotsb + \lambda_n d_n) \leq c\). Therefore, this inequality is true as well.
Exercise 4.17
Exercise 4.18
To show the power function is a convex function over the interval \((a, b)\) where \(a \geq 0\), we can show that its second derivative is always greater than or equal to 0. If \(f(x) = x^p\), then \(f''(x) = p(p - 1) \cdot x^{p - 2}\). Since \(p \geq 1\), \(p(p - 1) \geq 0\). Moreover, \(x^{p - 2} > 0\) for all \(x \in (a, b)\) and the following proves this inequality. Since \(p - 2\) is a rational number, we can write \(p - 2 = a/b\) and so \(\sqrt[b]{x^a}\). Since \(x > 0\), \(x^a > 0\) and taking the \(b\)-th root of a positive number will also be positive so \(x^{p - 2} > 0\). Since \(p(p - 1) \geq 0\) and \(x^{p - 2} > 0\), this means \(p(p - 1) x^{p - 2} \geq 0\) and so the power function is convex.