Exercises: Derivatives 1 – "Math for Non-Geeks"

verweis=Math for Non-Geeks

Computing derivatives with differential quotients

Exercises: derivative and differentiability

Exercise (Differentiable power function)

Show that the power function $f : ℝ ∖ {0}, f (x) = x^{- 2} = \frac{1}{x^{2}}$ is differentiable at $ξ = - 2$ and compute the derivative. What is the derivative of $f$ at any $\tilde{x} \in ℝ ∖ {0}$ ?

Solution (Differentiable power function)

The differential quotient of $f$ at $ξ = - 2$ is given by

$\begin{matrix} f^{'} (- 2) & = lim_{h \to 0} \frac{f (- 2 + h) - f (- 2)}{h} = lim_{h \to 0} \frac{\frac{1}{(- 2 + h)^{2}} - \frac{1}{(- 2)^{2}}}{h} \\ = lim_{h \to 0} \frac{\frac{(- 2)^{2} - (- 2 + h)^{2}}{(- 2)^{2} \cdot (- 2 + h)^{2}}}{h} = lim_{h \to 0} \frac{4 - (4 - 4 h + h^{2})}{4 h (h^{2} - 4 h + 4)} \\ = lim_{h \to 0} \frac{4 h - h^{2}}{4 h^{3} - 16 h^{2} + 16 h} = lim_{h \to 0} \frac{4 - h}{4 h^{2} - 16 h + 16} \\ = \frac{4 - 0}{0 + 0 + 16} = \frac{4}{16} = \frac{1}{4} \end{matrix}$

So $f$ is differentiable at $ξ = - 2$ , with derivative $f^{'} (- 2) = \frac{1}{4}$ . For a general $\tilde{x} \in ℝ ∖ {0}$ there is

$\begin{matrix} f^{'} (\tilde{x}) & = lim_{h \to 0} \frac{f (\tilde{x} + h) - f (\tilde{x})}{h} = lim_{h \to 0} \frac{\frac{1}{(\tilde{x} + h)^{2}} - \frac{1}{{\tilde{x}}^{2}}}{h} \\ = lim_{h \to 0} \frac{\frac{{\tilde{x}}^{2} - (\tilde{x} + h)^{2}}{{\tilde{x}}^{2} \cdot (\tilde{x} + h)^{2}}}{h} = lim_{h \to 0} \frac{{\tilde{x}}^{2} - ({\tilde{x}}^{2} + 2 \tilde{x} h + h^{2})}{{\tilde{x}}^{2} h ({\tilde{x}}^{2} + 2 \tilde{x} h + h^{2})} \\ = lim_{h \to 0} \frac{- 2 \tilde{x} h - h^{2}}{{\tilde{x}}^{4} h + 2 {\tilde{x}}^{3} h^{2} + {\tilde{x}}^{2} h^{3}} = lim_{h \to 0} \frac{- 2 \tilde{x} - h}{{\tilde{x}}^{4} + 2 {\tilde{x}}^{3} h + {\tilde{x}}^{2} h^{2}} \\ = \frac{- 2 \tilde{x} - 0}{{\tilde{x}}^{4} + 0 + 0} = - \frac{2}{{\tilde{x}}^{3}} \end{matrix}$

Exercise (Derivative of a product function)

Let $f : ℝ \to ℝ$ defined by

$f (x) = x (x - 1) (x - 2) \cdot \dots \cdot (x - 1000) = \prod_{k = 0}^{1000} (x - k)$

Determine $f^{'} (0)$ .

Solution (Derivative of a product function)

There is

$\begin{matrix} f^{'} (0) = lim_{x \to 0} \frac{f (x) - f (0)}{x - 0} = lim_{x \to 0} \frac{\prod_{k = 0}^{1000} (x - k) - 0}{x - 0} = lim_{x \to 0} \frac{x \cdot \prod_{k = 1}^{1000} (x - k)}{x} \\ = lim_{x \to 0} \prod_{k = 1}^{1000} (x - k) \overset{(*)}{=} \prod_{k = 1}^{1000} (- k) = \underset{= 1}{\underset{⏟}{(- 1)^{1000}}} \cdot \underset{= 1000!}{\underset{⏟}{\prod_{k = 1}^{1000} k}} = 1000! \end{matrix}$

For $(*)$ we have used that $x \mapsto \prod_{k = 1}^{1000} (x - k)$ is continuous as a product of the continuous functions $x \mapsto x - k$ for $1 \leq k \leq 1000$ .

Exercise (Derivative of a function with case distinction)

Check whether the following functions are differentiable at $x = 0$ .

$f : ℝ \to ℝ, f (x) = {\begin{matrix} \cos (\frac{1}{x}) & for x \neq 0 \\ 0 & for x = 0 \end{matrix}$
$g : ℝ \to ℝ, g (x) = {\begin{matrix} x \cos (\frac{1}{x}) & for x \neq 0 \\ 0 & for x = 0 \end{matrix}$

Solution (Derivative of a function with case distinction)

Part 1: Since $\cos (\frac{1}{x})$ oscillates very quickly between $- 1$ and $- 1$ , just like $\sin (\frac{1}{x})$ for $x \to 0$ , it is to be expected that $f$ at $x = 0$ is not continuous. For this purpose we consider the null sequences $(a_{n})_{n \in ℕ} = (\frac{1}{2 n π})_{n \in ℕ}$ and $(b_{n})_{n \in ℕ} = (\frac{1}{(2 n - 1) π})_{n \in ℕ}$ . For these sequences

$f (a_{n}) = \cos (2 n π) = 1$

and

$f (b_{n}) = \cos ((2 n - 1) π) = - 1$

So $lim_{x \to 0} f (x)$ does not exist. According to the sequence criterion $f$ is therefore not continuous at zero and thus not differentiable.

Part 2: The function $g$ is continuous at zero by the sequence criterion, as $lim_{x \to 0} g (x) = 0$ . So we can consider the differential quotient:

$lim_{x \to 0} \frac{g (x) - g (0)}{x - 0} = lim_{x \to 0} \frac{x \cos (\frac{1}{x}) - 0}{x} = lim_{x \to 0} \cos (\frac{1}{x})$

In Part 1 we have shown that this limit value does not exist. Therefore $g$ is also not differentiable at zero.

Exercise (Criterion for non-differentiability of a general function at zero)

Let $f : (- 1, 1) \to ℝ$ . Show that: If $| f (x) | \geq | x |^{α}$ for some $0 < α < 1$ and $f (0) = 0$ , then $f$ is not differentiable at zero.

Solution (Criterion for non-differentiability of a general function at zero)

There is

$| \frac{f (x) - f (0)}{x - 0} | = \frac{| f (x) |}{| x |} \overset{| f (x) | \geq | x |^{α}}{\geq} \frac{| x |^{α}}{| x |} = \frac{1}{| x |^{1 - α}} \overset{x \to 0}{⟶} \infty$

So $f^{'} (0) = lim_{x \to 0} \frac{f (x) - f (0)}{x - 0}$ does not exist.

Exercise (Determining limits with the differential quotient)

Let $f : D \to ℝ$ be differentiable at $a \in D$ . Show that the following limits exist

$lim_{h \to 0} \frac{f (a + h^{2}) - f (a)}{h} = 0$
$lim_{h \to 0} \frac{f (a + c h) - f (a - d h)}{h} = (c + d) f^{'} (a)$ for $c, d \in ℝ$
$lim_{x \to a} \frac{x f (a) - a f (x)}{x - a} = f (a) - a f^{'} (a)$

How to get to the proof? (Determining limits with the differential quotient)

Since $f$ in $a$ is differentiable, there is

$lim_{h \to 0} \frac{f (a + h) - f (a)}{h} = f^{'} (a)$

In addition, from the exercises within the article "derivatives", we know that

$lim_{h \to 0} \frac{f (a - h) - f (a)}{h} = - f^{'} (x)$

The idea is to transform the limits so that we can calculate them using the differential quotient.

Solution (Determining limits with the differential quotient)

Part 1: Because $lim_{h \to 0} \frac{f (a + h) - f (a)}{h} = f^{'} (a)$ there is also $lim_{h \to 0} \frac{f (a + h^{2}) - f (a)}{h^{2}} = f^{'} (a)$ .

So

$lim_{h \to 0} \frac{f (a + h^{2}) - f (a)}{h} = lim_{h \to 0} h \cdot \frac{f (a + h^{2}) - f (a)}{h^{2}} = \underset{= 0}{\underset{⏟}{lim_{h \to 0} h}} \cdot \underset{= f^{'} (a)}{\underset{⏟}{lim_{h \to 0} \frac{f (a + h^{2}) - f (a)}{h^{2}}}} = 0$

Part 2: With $lim_{h \to 0} \frac{f (a + h) - f (a)}{h} = f^{'} (a)$ and $lim_{h \to 0} \frac{f (a - h) - f (a)}{h} = - f^{'} (a)$ there is also $lim_{h \to 0} \frac{f (a + c h) - f (a)}{c h} = f^{'} (a)$ and $lim_{h \to 0} \frac{f (a - d h) - f (a)}{d h} = - f^{'} (a)$ . Hence,

$\begin{matrix} lim_{h \to 0} \frac{f (a + c h) - f (a - d h)}{h} & = lim_{h \to 0} \frac{f (a + c h) - f (a) - (f (a - d h) - f (a))}{h} = lim_{h \to 0} \frac{f (a + c h) - f (a)}{h} - lim_{h \to 0} \frac{f (a - d h) - f (a)}{h} \\ = c \cdot \underset{= f^{'} (a)}{\underset{⏟}{lim_{h \to 0} \frac{f (a + c h) - f (a)}{c h}}} - d \cdot \underset{= - f^{'} (a)}{\underset{⏟}{lim_{h \to 0} \frac{f (a - d h) - f (a)}{d h}}} = c f^{'} (a) + d f^{'} (a) = (c + d) f^{'} (a) \end{matrix}$

Part 3: Here we need the "original" differential quotient $lim_{x \to a} \frac{f (x) - f (a)}{x - a} = f^{'} (a)$ :

$\begin{matrix} lim_{x \to a} \frac{x f (a) - a f (x)}{x - a} & = lim_{x \to a} \frac{x f (a) - a f (a) - (a f (x) - a f (a))}{x - a} = lim_{x \to a} \frac{x f (a) - a f (a)}{x - a} - lim_{x \to a} \frac{a f (x) - a f (a)}{x - a} \\ = lim_{x \to a} \frac{(x - a) f (a)}{x - a} - lim_{x \to a} \frac{a (f (x) - f (a))}{x - a} = c \cdot \underset{= f (a)}{\underset{⏟}{lim_{x \to a} f (a)}} - a \cdot \underset{= f^{'} (a)}{\underset{⏟}{lim_{x \to a} \frac{f (x) - f (a)}{x - a}}} = f (a) - a f^{'} (a) \end{matrix}$

Exercise (Implication of differentiability)

Let $f : ℝ \to ℝ$ be differentiable at $\tilde{x}$ . Further, let $(a_{n})_{n \in ℕ}$ and $(b_{n})_{n \in ℕ}$ be sequences with $a_{n} < \tilde{x} < b_{n}$ for all $n \in ℕ$ , as well as $lim_{n \to \infty} (a_{n} - b_{n}) = 0$ . Show that then, there is

$lim_{n \to \infty} \frac{f (a_{n}) - f (b_{n})}{a_{n} - b_{n}} = f^{'} (\tilde{x})$

Additional question: Does the converse statement also hold? I.e. does the limit value $lim_{n \to \infty} \frac{f (a_{n}) - f (b_{n})}{a_{n} - b_{n}}$ with sequences $(a_{n})$ and $(b_{n})$ as above exist, so $f$ is differentiable at $\tilde{x}$ , and $f^{'} (\tilde{x})$ is equal to this limit?

Hint: Show first that $\frac{f (a_{n}) - f (b_{n})}{a_{n} - b_{n}} = \frac{f (a_{n}) - f (\tilde{x})}{a_{n} - \tilde{x}} + \frac{a_{n} - \tilde{x}}{a_{n} - b_{n}} (\frac{f (a_{n}) - f (\tilde{x})}{a_{n} - \tilde{x}} - \frac{f (b_{n}) - f (\tilde{x})}{b_{n} - \tilde{x}})$

Solution (Implication of differentiability)

There is

$\begin{matrix} \frac{f (a_{n}) - f (b_{n})}{a_{n} - b_{n}} & = \frac{f (a_{n}) - f (\tilde{x}) + f (\tilde{x}) - f (b_{n})}{a_{n} - b_{n}} \\ = \frac{f (a_{n}) - f (\tilde{x})}{a_{n} - b_{n}} + \frac{f (\tilde{x}) - f (b_{n})}{a_{n} - b_{n}} \\ = \frac{f (a_{n}) - f (\tilde{x})}{a_{n} - \tilde{x}} + \frac{f (a_{n}) - f (\tilde{x})}{a_{n} - b_{n}} - \frac{f (a_{n}) - f (\tilde{x})}{a_{n} - \tilde{x}} - \frac{f (b_{n}) - f (\tilde{x})}{a_{n} - b_{n}} \\ = \frac{f (a_{n}) - f (\tilde{x})}{a_{n} - \tilde{x}} + \frac{(a_{n} - \tilde{x} - (a_{n} - b_{n})) (f (a_{n}) - f (\tilde{x})}{(a_{n} - b_{n}) (a_{n} - \tilde{x})} - \frac{f (b_{n}) - f (\tilde{x})}{a_{n} - b_{n}} \\ = \frac{f (a_{n}) - f (\tilde{x})}{a_{n} - \tilde{x}} + \frac{b_{n} - \tilde{x}}{a_{n} - b_{n}} \cdot \frac{f (a_{n}) - f (\tilde{x})}{a_{n} - \tilde{x}} - \frac{b_{n} - \tilde{x}}{a_{n} - b_{n}} \cdot \frac{f (b_{n}) - f (\tilde{x})}{b_{n} - \tilde{x}} \\ = \underset{\to f^{'} (\tilde{x})}{\underset{⏟}{\frac{f (a_{n}) - f (\tilde{x})}{a_{n} - \tilde{x}}}} + \underset{| \dots | \leq 1}{\underset{⏟}{\frac{a_{n} - \tilde{x}}{a_{n} - b_{n}}}} \cdot \underset{\to f^{'} (\tilde{x}) - f^{'} (\tilde{x}) = 0}{\underset{⏟}{(\frac{f (a_{n}) - f (\tilde{x})}{a_{n} - \tilde{x}} - \frac{f (b_{n}) - f (\tilde{x})}{b_{n} - \tilde{x}})}} \end{matrix}$

Since now the product of a bounded sequence and a null sequence converges to zero, there is with the calculation rules for sequences

$lim_{n \to \infty} \frac{f (a_{n}) - f (b_{n})}{a_{n} - b_{n}} = f^{'} (\tilde{x}) + 0 = f^{'} (\tilde{x})$

Concerning the additional question: The converse is false. Let us consider the following function, which is not continuous at $\tilde{x} = 0$ (and therefore not differentiable):

$f : ℝ \to ℝ, f (x) = {\begin{matrix} 1 & for x \neq 0, \\ 0 & for x = 0 \end{matrix}$

Then, there is for all null sequences $(a_{n})$ and $(b_{n})$ with $a_{n} < 0 < b_{n}$ :

$lim_{n \to \infty} \frac{f (a_{n}) - f (b_{n})}{a_{n} - b_{n}} = lim_{n \to \infty} \frac{1 - 1}{a_{n} - b_{n}} = lim_{n \to \infty} \frac{0}{a_{n} - b_{n}} = 0$

Exercises: examples for derivatives

Exercise (Derivatives of linear and quadratic functions)

Determine (using the definition) the derivative of a linear function

$f : ℝ \to ℝ, f (x) = a x + b$

and of a quadratic function

$g : ℝ \to ℝ, g (x) = a x^{2} + b x + c$

with $a, b, c \in ℝ$ .

Solution (Derivatives of linear and quadratic functions)

1. linear function: For $\tilde{x} \in ℝ$ there is

$f^{'} (\tilde{x}) = lim_{x \to \tilde{x}} \frac{f (x) - f (\tilde{x})}{x - \tilde{x}} = lim_{x \to \tilde{x}} \frac{a x + b - (a \tilde{x} + b)}{x - \tilde{x}} = lim_{x \to \tilde{x}} \frac{a x + b - a \tilde{x} - b}{x - \tilde{x}} = lim_{x \to \tilde{x}} \frac{a x - a \tilde{x}}{x - \tilde{x}} = lim_{x \to \tilde{x}} \frac{a (x - \tilde{x})}{x - \tilde{x}} = lim_{x \to \tilde{x}} a = a$

2. quadratic function: For $\tilde{x} \in ℝ$ there is

$\begin{matrix} g^{'} (\tilde{x}) & = lim_{x \to \tilde{x}} \frac{g (x) - g (\tilde{x})}{x - \tilde{x}} = lim_{x \to \tilde{x}} \frac{a x^{2} + b x + c - (a {\tilde{x}}^{2} + b \tilde{x} + c)}{x - \tilde{x}} = lim_{x \to \tilde{x}} \frac{a x^{2} + b x - a {\tilde{x}}^{2} - b \tilde{x}}{x - \tilde{x}} \\ = lim_{x \to \tilde{x}} \frac{a (x^{2} - {\tilde{x}}^{2}) + b (x - \tilde{x})}{x - \tilde{x}} = lim_{x \to \tilde{x}} \frac{a (x + \tilde{x}) (x - \tilde{x})}{x - \tilde{x}} + lim_{x \to \tilde{x}} \frac{b (x - \tilde{x})}{x - \tilde{x}} = lim_{x \to \tilde{x}} a (x + \tilde{x}) + lim_{x \to \tilde{x}} b = 2 a \tilde{x} + b \end{matrix}$

Exercise (Derivative of the logarithm function)

Compute the derivative of the natural logarithm function

$f : ℝ^{+} \to ℝ, f (x) = \ln x$

directly, using the differential quotient.

Solution (Derivative of the logarithm function)

1st way:

For $\tilde{x} \in ℝ^{+}$ there is

$f^{'} (\tilde{x}) = lim_{x \to \tilde{x}} \frac{f (x) - f (\tilde{x})}{x - \tilde{x}} = lim_{x \to \tilde{x}} \frac{\ln x - \ln \tilde{x}}{x - \tilde{x}}$

Now for $0 < x < \tilde{x}$ we have the inequality

$\frac{1}{\tilde{x}} \leq \frac{f (x) - f (\tilde{x})}{x - \tilde{x}} \leq \frac{1}{x}$

If we swap the roles of $x$ and $\tilde{x}$ , then there is

$\frac{1}{x} \leq \frac{f (\tilde{x}) - f (x)}{\tilde{x} - x} \leq \frac{1}{\tilde{x}}$

Since the left and right-hand sides of the inequality for $x \to \tilde{x}$ converge towards $\frac{1}{\tilde{x}}$ , the squeeze theorem implies

$f^{'} (\tilde{x}) = lim_{x \to \tilde{x}} \frac{\ln x - \ln \tilde{x}}{x - \tilde{x}} = \frac{1}{\tilde{x}}$

2nd way: $h$ -method

For $\tilde{x} \in ℝ^{+}$ there is

$\begin{matrix} f^{'} (\tilde{x}) & = lim_{h \to \tilde{0}} \frac{f (\tilde{x} + h) - f (\tilde{x})}{h} = lim_{h \to 0} \frac{\ln (\tilde{x} + h) - \ln (\tilde{x})}{h} = lim_{h \to 0} \frac{\ln (\frac{\tilde{x} + h}{\tilde{x}})}{h} = lim_{h \to 0} \frac{\ln (1 + \frac{h}{\tilde{x}})}{h} = lim_{h \to 0} \frac{1}{\tilde{x}} \frac{\ln (1 + \frac{h}{\tilde{x}})}{\frac{h}{\tilde{x}}} \\ \overset{y = \frac{h}{\tilde{x}}}{=} lim_{y \to 0} \frac{1}{\tilde{x}} \frac{\ln (1 + y)}{y} = \frac{1}{\tilde{x}} \underset{= 1}{\underset{⏟}{lim_{y \to 0} \frac{\ln (1 + y)}{y}}} = \frac{1}{\tilde{x}} \end{matrix}$

Exercise (Computing the derivatives of hyperbolic functions

\sinh

\cosh

and

\tanh

)

Determine die derivatives of the following functions using the differential quotient

$\sinh : ℝ \to ℝ, \sinh (x) := \frac{e^{x} - e^{- x}}{2}$
$\cosh : ℝ \to ℝ, \cosh (x) := \frac{e^{x} + e^{- x}}{2}$
$\tanh : ℝ \to ℝ, \tanh (x) := \frac{\sinh}{\cosh}$

Solution (Computing the derivatives of hyperbolic functions

\sinh

\cosh

and

\tanh

)

Part 1: Let $x \in ℝ$ . Then, there is

$\begin{matrix} \sinh^{'} (x) = lim_{h \to 0} \frac{\sinh (x + h) - \sinh (x)}{h} = & lim_{h \to 0} \frac{\frac{e^{x + h} - e^{- (x + h)}}{2} - \frac{e^{x} - e^{- x}}{2}}{h} \\ = & lim_{h \to 0} \frac{e^{x} (e^{h} - 1) - e^{- x} (e^{- h} - 1)}{2 h} \\ = & lim_{h \to 0} \frac{e^{x} (e^{h} - 1)}{2 h} - lim_{h \to 0} \frac{e^{- x} (e^{- h} - 1)}{2 h} \\ ↓ substitution (2nd limit) \tilde{h} = - h \\ = & lim_{h \to 0} \frac{e^{x} (e^{h} - 1)}{2 h} - lim_{\tilde{h} \to 0} \frac{e^{- x} (e^{\tilde{h}} - 1)}{- 2 \tilde{h}} \\ = & \frac{e^{x}}{2} lim_{h \to 0} \frac{(e^{h} - 1)}{h} + \frac{e^{- x}}{2} lim_{\tilde{h} \to 0} \frac{(e^{\tilde{h}} - 1)}{\tilde{h}} \\ ↓ lim_{h \to 0} \frac{(e^{h} - 1)}{h} = 1 \\ = & \frac{e^{x} + e^{- x}}{2} \\ = & \cosh (x) \end{matrix}$

Alternative proof:

$\begin{matrix} \sinh^{'} (x) & = lim_{h \to 0} \frac{\sinh (x + h) - \sinh (x)}{h} \\ ↓ addition theorem: \sinh (x + y) = \sinh (x) \cosh (y) + \cosh (x) \sinh (y) \\ = lim_{h \to 0} \frac{\sinh (x) \cosh (h) + \cosh (x) \sinh (h) - \sinh (x)}{h} \\ = \sinh (x) \cdot \underset{= 0}{\underset{⏟}{lim_{h \to 0} \frac{\cosh (h) - 1}{h}}} + \cosh (x) \cdot \underset{= 1}{\underset{⏟}{lim_{h \to 0} \frac{\sinh (h)}{h}}} \\ = \cosh (x) \end{matrix}$

Part 2: Let $x \in ℝ$ . Then, there is

$\begin{matrix} \cosh^{'} (x) = lim_{h \to 0} \frac{\cosh (x + h) - \cosh (x)}{h} \\ = & lim_{h \to 0} \frac{\frac{e^{x + h} + e^{- (x + h)}}{2} - \frac{e^{x} + e^{- x}}{2}}{h} \\ = & lim_{h \to 0} \frac{e^{x} (e^{h} - 1) + e^{- x} (e^{- h} - 1)}{2 h} \\ = & lim_{h \to 0} \frac{e^{x} (e^{h} - 1)}{2 h} + lim_{h \to 0} \frac{e^{- x} (e^{- h} - 1)}{2 h} \\ ↓ substitution (2nd limit) \tilde{h} = - h \\ = & lim_{h \to 0} \frac{e^{x} (e^{h} - 1)}{2 h} + lim_{\tilde{h} \to 0} \frac{e^{- x} (e^{\tilde{h}} - 1)}{- 2 \tilde{h}} \\ = & \frac{e^{x}}{2} lim_{h \to 0} \frac{(e^{h} - 1)}{h} - \frac{e^{- x}}{2} lim_{\tilde{h} \to 0} \frac{(e^{\tilde{h}} - 1)}{\tilde{h}} \\ ↓ lim_{h \to 0} \frac{(e^{h} - 1)}{h} = 1 \\ = & \frac{e^{x} - e^{- x}}{2} \\ = & \sinh (x) \end{matrix}$

Alternative proof:

$\begin{matrix} \cosh^{'} (x) & = lim_{h \to 0} \frac{\cosh (x + h) - \cosh (x)}{h} \\ ↓ addition theorem: \cosh (x + y) = \cosh (x) \cosh (y) + \sinh (x) \sinh (y) \\ = lim_{h \to 0} \frac{\cosh (x) \cosh (h) + \sinh (x) \sinh (h) - \cosh (x)}{h} \\ = \cosh (x) \cdot \underset{= 0}{\underset{⏟}{lim_{h \to 0} \frac{\cosh (h) - 1}{h}}} + \sinh (x) \cdot \underset{= 1}{\underset{⏟}{lim_{h \to 0} \frac{\sinh (h)}{h}}} \\ = \sinh (x) \end{matrix}$

Part 3: Let $x \in ℝ$ . Then, there is

$\tanh (x) = \frac{\sinh (x)}{\cosh (x)} = \frac{\frac{e^{x} - e^{- x}}{2}}{\frac{e^{x} + e^{- x}}{2}} = \frac{2 (e^{x} - e^{- x})}{2 (e^{x} + e^{- x})} = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$

So

$\begin{matrix} \tanh^{'} (x) = lim_{h \to 0} \frac{\tanh (x + h) - \tanh (x)}{h} \\ = & lim_{h \to 0} \frac{\frac{e^{x + h} - e^{- (x + h)}}{e^{x + h} + e^{- (x + h)}} - \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}}{h} \\ = & lim_{h \to 0} \frac{(e^{x + h} - e^{- (x + h)}) (e^{x} + e^{- x}) - (e^{x} - e^{- x}) (e^{x + h} + e^{- (x + h)})}{h (e^{x + h} + e^{- (x + h)}) (e^{x} + e^{- x})} \\ = & lim_{h \to 0} \frac{e^{x + h} e^{x} - e^{- x - h} e^{x} + e^{- x} e^{x + h} - e^{- x} e^{- x - h} - e^{x} e^{x + h} - e^{x} e^{- x - h} + e^{- x} e^{x + h} + e^{- x} e^{- x - h}}{h (e^{x + h} + e^{- (x + h)}) (e^{x} + e^{- x})} \\ = & lim_{h \to 0} \frac{2 e^{- x} e^{x + h} - 2 e^{x} e^{- x - h}}{h (e^{x + h} + e^{- (x + h)}) (e^{x} + e^{- x})} \\ = & lim_{h \to 0} \frac{2 e^{h} - 2 e^{- h}}{h (e^{x + h} + e^{- (x + h)}) (e^{x} + e^{- x})} \\ = & lim_{h \to 0} \frac{2 e^{- h} (e^{2 h} - 1)}{h (e^{x + h} + e^{- (x + h)}) (e^{x} + e^{- x})} \\ = & lim_{h \to 0} 2 e^{- h} \cdot \frac{(e^{2 h} - 1)}{h} \cdot \frac{1}{(e^{x + h} + e^{- (x + h)}) (e^{x} + e^{- x})} \\ = & lim_{h \to 0} 2 e^{- h} \cdot 2 \frac{(e^{2 h} - 1)}{2 h} \cdot \frac{1}{(e^{x + h} + e^{- (x + h)}) (e^{x} + e^{- x})} \\ = & lim_{h \to 0} 2 e^{- h} \cdot lim_{h \to 0} 2 \frac{(e^{2 h} - 1)}{2 h} \cdot lim_{h \to 0} 2 e^{- h} \frac{1}{(e^{x + h} + e^{- (x + h)}) (e^{x} + e^{- x})} \\ ↓ lim_{h \to 0} \frac{(e^{2 h} - 1)}{2 h} = 1 \\ = & 2 e^{0} \cdot 2 \cdot \frac{1}{(e^{x} + e^{- x}) (e^{x} + e^{- x})} \\ = & \frac{4}{(e^{x} + e^{- x})^{2}} \\ = & \frac{1}{{(\frac{e^{x} + e^{- x}}{2})}^{2}} \\ = & \frac{1}{\cosh^{2} (x)} \end{matrix}$

Alternative proof:

$\begin{matrix} \tanh^{'} (x) & = lim_{h \to 0} \frac{\tanh (x + h) - \tanh (x)}{h} \\ = lim_{h \to 0} \frac{\frac{\sinh (x + h)}{\cosh (x + h)} - \frac{\sinh (x)}{\cosh (x)}}{h} \\ = lim_{h \to 0} \frac{\frac{\sinh (x + h) \cosh (x) - \sinh (x) \cosh (x + h)}{\cosh (x + h) \cosh (x)}}{h} \\ = lim_{h \to 0} \frac{\sinh (x + h) \cosh (x) - \sinh (x) \cosh (x + h)}{h \cosh (x + h) \cosh (x)} \\ ↓ addition theorems: \sinh (x + y) = \sinh (x) \cosh (y) + \cosh (x) \sinh (y) and \cosh (x + y) = \cosh (x) \cosh (y) + \sinh (x) \sinh (y) \\ = lim_{h \to 0} \frac{(\sinh (x) \cosh (h) + \cosh (x) \sinh (h)) \cosh (x) - \sinh (x) (\cosh (x) \cosh (h) + \sinh (x) \sinh (h))}{h \cosh (x + h) \cosh (x)} \\ = lim_{h \to 0} \frac{\sinh (x) \cosh (x) \cosh (h) + \cosh^{2} (x) \sinh (h) - \sinh (x) \cosh (x) \cosh (h) - \sinh^{2} (x) \sinh (h)}{h \cosh (x + h) \cosh (x)} \\ = lim_{h \to 0} \frac{\cosh^{2} (x) \sinh (h) - \sinh^{2} (x) \sinh (h)}{h \cosh (x + h) \cosh (x)} \\ = lim_{h \to 0} \frac{(\cosh^{2} (x) - \sinh^{2} (x)) \sinh (h)}{h \cosh (x + h) \cosh (x)} \\ ↓ \cosh^{2} (x) - \sinh^{2} (x) = 1 \\ = lim_{h \to 0} \frac{\sinh (h)}{h \cosh (x + h) \cosh (x)} \\ = lim_{h \to 0} \frac{\sinh (h)}{h} \cdot \frac{1}{\cosh (x + h) \cosh (x)} \\ ↓ lim_{h \to 0} \frac{\sinh (h)}{h} = 1 \\ = \frac{1}{\cosh^{2} (x)} \end{matrix}$

Computation rules for derivatives

Applying the computation rules

Exercise (derivatives of a power function)

Show by induction in $n \in ℕ$ , that the power function

$f : ℝ \to ℝ, f (x) = x^{n}$

is differentiable with

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

Proof (derivatives of a power function)

Induction base: $n = 1$

If $f (x) = x^{1} = x$ , then there is $f^{'} (x) = 1 = 1 \cdot x^{0}$

Induction assumption:

For $f (x) = x^{n}$ with $n \in ℕ$ , there is

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

Induction step: $n \to n + 1$

Let $f (x) = x^{n + 1} = x \cdot x^{n}$ . Then $f$ is differentiable by induction assumption and the product rule. For $x \in ℝ$ there is $f^{'} (x) \overset{IV}{=} 1 \cdot x^{n} + x \cdot n x^{n - 1} = x^{n} + n x^{n} = (n + 1) x^{n}$

Exercise (Derivatives of secant and cosecant)

The functions $\sec$ (secant) and $\csc$ (cosecant) are defined as follows:

$\sec (x) = (\cos x)^{- 1} = \frac{1}{\cos x}$

as well as

$\csc (x) = (\sin x)^{- 1} = \frac{1}{\sin x}$

Determine their domain of definition and all derivatives.

Solution (Derivatives of secant and cosecant)

Part 1: secant

domain of definition: $\sec$ is well-defined $⟺$ $\cos x \neq 0$ $⟺$ $x \neq \frac{π}{2} + k π$ $\forall k \in ℤ$ $⟹$ $D_{1} = ℝ ∖ {\frac{π}{2} + k π ∣ k \in ℤ}$

Derivative: Since $\cos$ is differentiable on $D_{1}$ , there is with the chain rule

$\sec (x) = (- 1) (\cos x)^{- 2} \cdot (- \sin x) = \frac{\sin x}{\cos^{2} x} = \sec x \cdot \tan x = \frac{\sec^{2} x}{\csc x}$

Part 2: cosecant

domain of definition: $\csc$ is well-defined $⟺$ $\sin x \neq 0$ $⟺$ $x \neq k π$ $\forall k \in ℤ$ $⟹$ $D_{2} = ℝ ∖ {k π ∣ k \in ℤ}$

Derivative: Since $\sin$ is differentiable on $D_{2}$ , there is with the chain rule

$\csc (x) = (- 1) (\sin x)^{- 2} \cdot (\cos x) = - \frac{\cos x}{\sin^{2} x} = \csc x \cdot \cot x = - \frac{\csc^{2} x}{\sec x}$

Exercise (Computing derivatives)

Determine the domain of definition of the following functions, as well as their derivatives

$f_{1} (x) = x^{2} e^{- x}$
$f_{2} (x) = \ln (\ln (\ln (x)))$
$f_{3} (x) = \sin (x \cos (x^{2}))$
$f_{4} (x) = \frac{1 + x^{2}}{\sqrt{x^{2} - 1}}$
$f_{5} (x) = x \cdot | x |$

Solution (Computing derivatives)

Part 1:

domain of definition: $ℝ$

Derivative: For $x \in ℝ$ there is with the product rule

${f_{1}}^{'} (x) = 2 x e^{- x} + x^{2} (- e^{- x}) = x e^{- x} (2 - x)$

Part 2:

domain of definition: $(e, \infty)$ , as

$\ln (\ln (x)) > 0 ⟺ \ln (x) > 1 ⟺ x > e$

Derivative: For $x \in (0, \infty)$ there is with the chain rule

${f_{2}}^{'} (x) = \frac{1}{\ln (\ln (x))} \cdot \frac{1}{\ln (x)} \cdot \frac{1}{x} = \frac{1}{x \ln (x) \ln (\ln (x))}$

Part 3:

domain of definition: $ℝ$

Derivative: For $x \in ℝ$ there is with the chain- and product rule

${f_{3}}^{'} (x) = \cos (x \cos (x^{2})) [1 \cdot \cos (x^{2}) + x (- \sin (x^{2}) \cdot 2 x)] = \cos (x \cos (x^{2})) [\cos (x^{2}) - 2 x^{2} \sin (x^{2})]$

Part 4:

domain of definition: $(- \infty, - 1) \cup (1, \infty)$ , since there must be

$\sqrt{x^{2} - 1} \neq 0 ⟺ x^{2} \neq 1 ⟺ x \neq \pm 1$
$x^{2} - 1 > 0 ⟺ x^{2} > 1 ⟺ | x | > 1 ⟺ (x < - 1) \lor (x > 1)$

Derivative: For $x \in ℝ$ there is with the quotient rule

${f_{4}}^{'} (x) = \frac{2 x \sqrt{x^{2} - 1} - (1 + x^{2}) \frac{2 x}{2 \sqrt{x^{2} - 1}}}{x^{2} - 1} = \frac{2 x (x^{2} - 1) - x (1 + x^{2})}{(x^{2} - 1)^{\frac{3}{2}}} = \frac{x (2 x^{2} - 2 - 1 - x^{2})}{(x^{2} - 1)^{\frac{3}{2}}} = \frac{x (x^{2} - 3)}{(x^{2} - 1)^{\frac{3}{2}}}$

Part 5:

domain of definition: $ℝ$

Derivative:

For $x < 0$ there is $f_{5} (x) = x^{2}$ $⟹$ ${f_{5}}^{'} (x) = 2 x$

For $x < 0$ there is $f_{5} (x) = - x^{2}$ $⟹$ ${f_{5}}^{'} (x) = - 2 x$

Further there is

$lim_{x \to 0 +} \frac{f_{5} (x) - f_{5} (0)}{x - 0} = lim_{x \to 0 +} \frac{x^{2} - 0}{x} = lim_{x \to 0 +} x = 0$

as well as

$lim_{x \to 0 -} \frac{f_{5} (x) - f_{5} (0)}{x - 0} = lim_{x \to 0 -} \frac{- x^{2} - 0}{x} = lim_{x \to 0 -} - x = 0$

So we have

${f_{5}}^{'} (0) = lim_{x \to 0} \frac{f_{5} (x) - f_{5} (0)}{x - 0} = 0$

Concluding all three cases, we get for $x \in ℝ$

${f_{5}}^{'} (x) = 2 | x |$

Exercise (derivatives of exponential functions)

Determine the derivatives of the following functions on their domain of definition ( $a > 0$ )

$f : ℝ^{+} \to ℝ, f (x) = x^{(x^{x})}$
$g : ℝ^{+} \to ℝ, g (x) = (x^{x})^{x}$
$h : ℝ^{+} \to ℝ, h (x) = x^{(x^{a})}$
$i : ℝ^{+} \to ℝ, i (x) = x^{(a^{x})}$
$j : ℝ^{+} \to ℝ, j (x) = a^{(x^{x})}$

For the function $g$ we may leave out the bracket, since in general $x^{x^{x}} = x^{(x^{x})}$ is well-defined.

Solution (derivatives of exponential functions)

Part 1: There is $g (x) = x^{(x^{x})} = \exp (x^{x} \ln x)$ . The function $x \mapsto x^{x} = \exp (x \ln (x))$ is differentiable with $(x^{x})^{'} = x^{x} (\ln (x) + 1)$ . Hence $g$ is differentiable by the chain- and product rule and for $x \in ℝ^{+}$ there is

$f^{'} (x) = \exp (x^{x} \ln x) [x^{x} (\ln x + 1) \cdot \ln x + x^{x} \cdot \frac{1}{x}] = \exp (x \ln x) [x^{x} (\ln x + 1) \ln (x) + \frac{1}{x})] = x^{x^{x} + x} (\ln^{2} x + \ln x + \frac{1}{x})$

Part 2: There is $g (x) = (x^{x})^{x} = x^{x \cdot x} = x^{x^{2}} = \exp (x^{2} \ln x)$ . Hence $h$ is differentiable by the chain- and product rule and for $x \in ℝ^{+}$ there is

$g^{'} (x) = \exp (x^{2} \ln x) [2 x \ln x + x^{2} \cdot \frac{1}{x}] = \exp (x^{2} \ln x) [2 x \ln x + x] = \exp (x^{2} \ln x) [x (2 \ln x + 1)] = x^{x^{2} + 1} (2 \ln x + 1)$

Part 3: There is $h (x) = x^{(x^{a})} = \exp (x^{a} \ln x)$ . Hence $h$ is differentiable by the chain- and product rule and for $x \in ℝ^{+}$ there is

$h^{'} (x) = \exp (x^{a} \ln x) [a x^{a - 1} \cdot \ln x + x^{a} \cdot \frac{1}{x}] = \exp (x \ln x) [x^{a - 1} (a \ln x + 1)] = x^{x^{a} + a - 1} (a \ln x + 1)$

Part 4: There is $i (x) = (x^{(a^{x})}) = \exp (a^{x} \ln x)$ . Hence $i$ is differentiable by the chain- and product rule and for $x \in ℝ^{+}$ there is

$i^{'} (x) = \exp (a^{x} \ln x) [a^{x} \ln a \ln x + a^{x} \cdot \frac{1}{x}] = x^{(a^{x})} a^{x} [\ln a \ln x + \frac{1}{x})]$

Part 5: There is $j (x) = a^{(x^{x})} = \exp (x^{x} \ln a)$ . The function $x \mapsto x^{x} = \exp (x \ln (x))$ is differentiable with $(x^{x})^{'} = x^{x} (\ln (x) + 1)$ . Hence $j$ is differentiable by the chain- and product rule and for $x \in ℝ^{+}$ there is

$j^{'} (x) = \exp (x^{x} \ln a) [x^{x} (\ln x + 1) \cdot \ln a] = a^{(x^{x})} x^{x} \ln a (\ln x + 1)$

Exercise (Proof of sum formulas using the derivative)

Proofs by means of binomial theorem (missing) that for all $n \in ℕ$ :

$\sum_{k = 1}^{n} k (\binom{n}{k}) = n 2^{n - 1}$
$\sum_{k = 1}^{n} (- 1)^{k - 1} k (\binom{n}{k}) = 0$

Use the binomial theorem and set $x = 1$ . Then take the derivative on both sides.

Proof (Proof of sum formulas using the derivative)

For $x = 1$ the binomial theorem reads

$\sum_{k = 0}^{n} (\binom{n}{k}) y^{k} = (1 + y)^{n}$

for $n \in ℕ$ and $y \in ℝ$ . Now the left-hand side of the equation is a polynomial $f (y)$ and the right-hand side is a power function $g (y)$ . Both sides are therefore differentiable on $ℝ$ with

$f^{'} (y) = \sum_{k = 0}^{n} (\binom{n}{k}) k y^{k - 1} = \sum_{k = 1}^{n} k (\binom{n}{k}) y^{k - 1}$

and

$g^{'} (y) = n (1 + y)^{n - 1}$

Since $f \equiv g$ there is also $f^{'} \equiv g^{'}$ . So in particular

$f (1) = g (1) ⟺ \sum_{k = 1}^{n} k (\binom{n}{k}) = n 2^{n - 1}$

and

$f (- 1) = g (- 1) ⟺ \sum_{k = 1}^{n} (- 1)^{k - 1} k (\binom{n}{k}) = 0$

Exercise (Logarithmic derivatives)

Determine the logarithmic derivatives of the following functions

$f (x) = \frac{1}{x}$
$g (x) = \sec (x) = \frac{1}{\sin (x)}$
$h (x) = a^{x} = \exp (x \ln (a))$ with $a \in ℝ^{+}$
$h (x) = x^{x} = \exp (x \ln (x))$

Proof of computational laws

Exercise (Alternative proof of the product rule)

Prove that for differentiable $f, g : D \to ℝ$ the product rule

$(f g)^{'} = f^{'} g + g^{'} f$

holds - by using the chain rule.

Hint: There is: $f g = \frac{1}{2} ((f + g)^{2} - f^{2} - g^{2})$

Proof (Alternative proof of the product rule)

The function $g : ℝ \to ℝ, g (x) = x^{2}$ is differentiable on $ℝ$ with

$g^{'} (x) = 2 x$

By der chain rule, we hence have that $g \circ f = f^{2} : D \to ℝ$ is differentiable with

$(g \circ f)^{'} (x) = (g \circ f)^{'} (x) \cdot f^{'} (x) = 2 f (x) f^{'} (x)$

for all $x \in D$ . Using the hint, we get with the factor- and sum rule

$\begin{matrix} (f g)^{'} (x) & = (\frac{1}{2} ((f + g)^{2} - f^{2} - g^{2}))^{'} (x) \\ = \frac{1}{2} [(2 (f (x) + g (x))) (f^{'} (x) + g^{'} (x)) - 2 f (x) f^{'} (x) - 2 g (x) g^{'} (x)] \\ = \frac{1}{2} [2 f (x) f^{'} (x) + 2 f (x) g^{'} (x) + 2 g (x) f^{'} (x) + 2 g (x) g^{'} (x) - 2 f (x) f^{'} (x) - 2 g (x) g^{'} (x)] \\ = \frac{1}{2} [2 f (x) g^{'} (x) + 2 g (x) f^{'} (x)] \\ = f (x) g^{'} (x) + g (x) f^{'} (x) \end{matrix}$

Exercise (Special case of the chain rule)

Derive a general derivative formula for the following function:

$f_{1}^{f_{2}} : (0, \infty) \to ℝ, (f_{1}^{f_{2}}) (x) = \exp (f_{2} (x) \ln (f_{1} (x)))$

If $f_{1}, f_{2} : (0, \infty) \to ℝ$ are differentiable.

Solution (Special case of the chain rule)

There is

$f_{1}^{f_{2}} : (0, \infty) \to ℝ, f_{1} (x)^{f_{2} (x)} = \exp (f_{2} (x) \cdot \ln (f_{1} (x))) = g (f (x))$

with $g (x) = \exp (x)$ and $f (x) = f_{2} (x) \cdot \ln (f_{1} (x))$ for all $x \in (0, \infty)$ . The function $f$ is differentiable by the product rule with $f^{'} (x) = {f_{2}}^{'} (x) \ln (f_{1} (x)) + f_{2} (x) \frac{{f_{1}}^{'} (x)}{f_{1} (x)}$

By the chain rule, also $f_{1}^{f_{2}} : (0, \infty) \to ℝ$ is differentiable, and there is

$(f_{1}^{f_{2}})^{'} (x) = g^{'} (f (x)) \cdot f^{'} (x) = \exp (f_{2} (x)) \cdot [{f_{2}}^{'} (x) \ln (f_{1} (x)) + f_{2} (x) \frac{{f_{1}}^{'} (x)}{f_{1} (x)}] = f_{1} (x)^{f_{2} (x)} [{f_{2}}^{'} (x) \ln (f_{1} (x)) + f_{2} (x) \frac{{f_{1}}^{'} (x)}{f_{1} (x)}]$

Theorem (Computational laws for logarithmic derivatives)

For two differentiable functions $f$ and $g$ without zeros there is

$L (\sqrt[k]{f}) = \frac{1}{k} L (f)$ for $k \in ℕ$ and $f > 0$
$L (f^{α}) = α L (f)$ for $α \in ℝ$ and $f > 0$
$L (f + g) = L (f) + L (1 + \frac{f}{g})$

Exercises 2 →

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