Exercise 5

Preliminary remark: In this rather extensive task, the derivatives of the sine, cosine, exponential and logarithm functions are to be proven. Being able to differentiate these functions is extremely important and an indispensable tool.

In my opinion, the proof of these derivation rules as an exercise task makes little sense, because hardly any readers from the target group of the textbook will be able to carry out these proofs on their own, unless they are already familiar with the proofs from school or university, google the proofs online or use relationships and limits from a reference book (as recommended in the task), which in turn are adopted without proof.

In this respect, it remains unclear what this exercise is intended to achieve, because then you could immediately ask the students to look up the derivation rules in a reference book.

There are several ways to proof these rules. At least one of them should be part of the textbook. We will stick to the classcial proof and use the definition of the derivative.

Derivative of \(\sin x\)

We begin with the definition of the derivative:

\[ \frac{\mathrm{d}}{\mathrm{d}x} \sin x = \lim_{h\to 0} \frac{\sin (x+h)- \sin x}{h} \]

Using the addition theorem for the sine, we get:

\[ = \lim_{h\to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h} \]

The fraction and the expression for the limit can be broken up. The terms without h can be pulled out and put in front of the limit expression:

\[ = \sin x \lim_{h\to 0} \frac{\cos h - 1}{h} + \cos x \lim_{h\to 0} \frac{\sin h}{h}\]

We’ll see in a minute, that the following is true for the limits:

\[ \lim_{h\to 0} \frac{\sin h}{h} = 1 \] \[ \lim_{h\to 0} \frac{\cos h - 1}{h} = 0 \]

We get: \[ \frac{\mathrm{d}}{\mathrm{d}x} \sin x = \cos x \]

Now let’s catch up on determining the two limit values, the tricky part of the proof.

At first, we look at \(\lim_{h\to 0} \frac{\sin h}{h}\). For \(h=0\) we get the undefined expression \(\frac{0}{0}\). But what happens for very small values of \(h\), in the vicinity of zero?

We will prove this geometrically by looking at the figure below. The arc BC is a quarter of the unit circle (a circle with radius 1). The area of the triangle AFE is calculated as “base length times height divided by 2”:

\[ A_{\Delta AFE} = \frac{1}{2} |\overline{AF}| \, |\overline{FE}| = \frac{1}{2}\cos \varphi \sin \varphi \]

The area of the sector ABE is given by

\[ A_{ABE} = \frac{\varphi}{2\pi} \pi r^2 = \frac{\varphi}{2}\]

(The area of the unit circle is \(\pi r^2 = \pi\), this is equivalent of the sector area
given by \(\varphi = 2\pi\). The sector with \(\varphi\) is the fractional part \(\frac{\varphi}{2\pi}\) of the circle area).

The area of the triangle ABD is given by \[ A_{\Delta ABD} = \frac{1}{2} |\overline{AB}| \, |\overline{BD}| = \frac{1}{2} \cdot 1 \cdot \tan \varphi \]

According to the picture, the following inequality holds true:

\[ A_{\Delta AFE} \le A_{ABE} \le A_{\Delta ABD} \] \[ \frac{1}{2}\cos \varphi \sin \varphi \le \frac{\varphi}{2} \le \frac{1}{2} \tan \varphi \]

Multiplication by 2 and division by \(\sin \varphi\) yields: \[ \cos \varphi \le \frac{\varphi}{\sin \varphi} \le \frac{1}{\cos \varphi} \]

The reciprocal of this inequality gives an estimation for the expression in question: \[ \frac{1}{\cos \varphi} \ge \frac{\sin \varphi}{\varphi} \ge \cos \varphi \]

\[\varphi \rightarrow 0 \Rightarrow \cos \varphi \rightarrow 1\] hence, \[1 \ge \lim_{\varphi \to 0} \frac{\sin \varphi }{\varphi} \ge 1 \Rightarrow \lim_{\varphi \to 0} \frac{\sin \varphi }{\varphi} = 1\]

Note

We’ve found an important approximation: \(\sin \varphi = \varphi\) for small values of \(\varphi\).

Now, let’s look at \(\lim_{h \to 0} \frac{\cos h - 1}{h}\). Again, we use a trick and the addition theorems for the cosine:

\[ \frac{\cos h -1 }{h} = \frac{\cos ( \frac{h}{2} + \frac{h}{2})-1}{h} = \frac{\cos \frac{h}{2} \, \cos \frac{h}{2} - \sin \frac{h}{2} \, \sin \frac{h}{2} -1}{h}\]

\[ = \frac{\cos^2 \frac{h}{2} - \sin^2 \frac{h}{2} -1}{h}\]

Using Pythagoras (\(\cos^2 x + \sin^2 x = 1\)) we get: \[ = \frac{1 - \sin^2 \frac{h}{2} - \sin^2 \frac{h}{2} -1}{h} = \frac{-2 \sin^2 \frac{h}{2}}{h} = - \sin (\frac{h}{2}) \frac {\sin \frac{h}{2}}{\frac{h}{2}}\]

\(\lim_{h\to 0} \frac{\sin h}{h}\) is already known to be 1. Hence, \[\lim_{h \to 0} \frac{\cos h - 1}{h} = - \lim_{h \to 0} \left( \sin (\frac{h}{2}) \frac {\sin \frac{h}{2}}{\frac{h}{2}} \right) = - \lim_{h \to 0} \sin (\frac{h}{2}) = 0\]

This completes our proof.

Derivative of \(\cos x\)

We’ll use the following equations: \[ \sin (x + \frac{\pi}{2}) = \cos x \] \[ \cos (x + \frac{\pi}{2}) = - \sin x\] You can verify these easily looking at the unit circle or using the addition theorems. \[ \frac{\mathrm{d}}{\mathrm{d}x} \cos x = \frac{\mathrm{d}}{\mathrm{d}x} \sin (x + \frac{\pi}{2}) = \cos (x + \frac{\pi}{2}) = - \sin x \]

Derivative of \(\mathrm{e}^x\)

As mentioned in the textbook, the value of Euler’s number \(\mathrm{e}\) is chosen so that the following applies: \[ \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{e}^x = \mathrm{e}^x\]

This equation is a definition for \(\mathrm{e}\), so there is nothing to proof here. But if you are interested in how to find the value of \(\mathrm{e}\) using this definition, then read on.

Again, we use the definition of the derivative as starting point: \[ \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{e}^x = \lim_{h \to 0} \frac{\mathrm{e}^{(x+h)}-\mathrm{e}^x}{h} = \lim_{h \to 0} \frac{\mathrm{e}^x \mathrm{e}^h -\mathrm{e}^x}{h} = \mathrm{e}^x \lim_{h \to 0} \frac{\mathrm{e}^h-1}{h} = \mathrm{e}^x \]

The value of \(\mathrm{e}\) must be chosen so that the following applies:
\[ \lim_{h \to 0} \frac{\mathrm{e}^h-1}{h} = 1 \]

Solving for \(\mathrm{e}^h\) yields: \[ \lim_{h \to 0} \mathrm{e}^h = \lim_{h \to 0} (1 + h) \Rightarrow \mathrm{e} = \lim_{h \to 0} \sqrt[h]{1+h} = \lim_{h \to 0} (1+h)^\frac{1}{h} \]

To simplify the calculation, we introduce the variable \(n := \frac{1}{h}\). \[ h \to 0 \Rightarrow n \to \infty\] \[ \mathrm{e} = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n \]

Using an electronic calculator we get an approximation value for \(\mathrm{e}\). For \(n = 10000\), we get \(\mathrm{e} = 2.718 \ldots\) with a precision of 3 digits.

Derivative of \(\ln x\)

Here is the trick: we start with the equation \[ x = \mathrm{e}^{\ln x} \]

since the natural logarithm is the inverse of the exponential function of base \(\mathrm{e}\).

Differentiating both sides of the equation yields (use the chain rule): \[ 1 = \mathrm{e}^{\ln x} \frac{\mathrm{d}}{\mathrm{d}x} \ln x = x \frac{\mathrm{d}}{\mathrm{d}x} \ln x \] \[ \Rightarrow \frac{\mathrm{d}}{\mathrm{d}x} \ln x = \frac{1}{x} \]