Other Functions and Their Derivatives

For the sake of developing technique, we mention several other functions and give their derivatives. Satisfactory definitions of the sine and cosine from the mathematical point of view will not be given in this course, but we shall present intuitive evidence for the formulas for their derivatives. The exponential and logarithmic functions will be discussed in more detail at a later stage in the course.

Function	Derivative
$\sin x$ ($x$ measured in radians)	$\cos x$
$\cos x$	$-\sin x$
$e^x$	$e^x$
$\log x$ ($={\log }_{e}x=\ln x$)	${\displaystyle \frac{1}{x}}$

Here $e$ is the number $2.71828\dots$, the base of the system of so-called natural logarithms. Whenever we write $\log x$, it will mean the logarithm to the base $e$. Now we can derive a rule for differentiating $a^x$, where $a$ is any fixed positive number. But first we must define $a^x$ for arbitrary exponent $x$. Assuming that the exponential and logarithm functions are defined, we define $a^x=e^{x\log a}$. Then if we set $f(u)=e^u$ (f'(u) = e^u also), $g(x)=x\log a$, we have $a^x=f(g(x))$, and

(a^x)' = f'(g(x)) \cdot g'(x) = e^{x \log a} \cdot \log a = a^x \log a .

Let us now form the difference quotients for $\sin x$ and $\cos x$. They are:

$$\frac{\sin (x+h)-\sin x}{h}=\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}$$

and

$$\frac{\cos (x+h)-\cos x}{h}=\frac{\cos x\cos h-\sin x\sin h-\cos x}{h},$$

or

\tag{1} \frac{\cos h - 1}{h} \cdot \sin x + \frac{\sin h}{h} \cdot \cos x ,

and

\tag{2} \frac{\cos h - 1}{h} \cdot \cos x - \frac{\sin h}{h} \cdot \sin x .

Thus we are led to consider

$$\underset{h\to 0}{lim}\frac{\cos h-1}{h}\text{and}\underset{h\to 0}{lim}\frac{\sin h}{h}.$$

We can reduce the first of these to the second as follows:

\begin{aligned} \frac{\cos h - 1}{h} &= \frac{\cos^2 h - 1}{h(\cos h + 1)} \\ &= \frac{-\sin^2 h}{h(\cos h + 1)} \\ & = \frac{\sin h}{h} \cdot \sin h \cdot \frac{-1}{\cos h + 1} . \end{aligned}

Of these three factors, the second goes to 0 as $h\to 0$, and the third approaches $-1/2$. If the limit of $\frac{\sin h}{h}$ exists at all, the limit of $\frac{\cos h-1}{h}$ will therefore be 0.

Now consider the diagram on the right. The area of the triangle OAB is less than or equal to that of the sector OCB, which in turn is less than or equal to that of triangle OCD. But we know:

$$\text{OAB}=\frac{1}{2}\sin h\cdot \cos h$$

sector OCB $={\displaystyle \frac{h}{2}}$ (since the area of the whole circle of radius 1 is $\pi$, and there are $2\pi$ radians in the circle.)

$$\text{OCD}=\frac{1}{2}\tan h=\frac{1}{2}\frac{\sin h}{\cos h}.$$

Thus

$$\frac{1}{2}\sin h\cdot \cos h\le \frac{h}{2}\le \frac{1}{2}\frac{\sin h}{\cos h}.$$

If , the first inequality gives

$$\frac{\sin h}{h}\le \frac{1}{\cos h},$$

and the second gives

$$\cos h\le \frac{\sin h}{h}.$$

Thus

$$\cos h\le \frac{\sin h}{h}\le \frac{1}{\cos h}.$$

But as $h\to 0$, $\cos h$ and $\frac{1}{\cos h}$ both approach 1. Since $\frac{\sin h}{h}$ lies between these quantities, it must also approach 1. A similar argument can be used for . Therefore we conclude that

$$\underset{h\to 0}{lim}\frac{\sin h}{h}=1,\underset{h\to 0}{lim}\frac{\cos h-1}{h}=0.$$

Taking limits in the difference quotients (1) and (2), we obtain the results originally given for the derivatives of the sine and cosine.

NOTE. Owing to the presence of undefined terms, you should regard this argument not as a proof, but rather as an effort to convince you that our statements are reasonable in the light of your experience.

Exercises