1 Differentiation

Physics 201 and 202 are calculus-based physics courses. This means that calculus will be used on a routine basis, most likely every day in class and on most homework assignments. It is very important that you be able to perform the basic calculus operations;
1. computations of limits and derivatives, including partial derivatives
2. expansions of functions, parameterization of curves
3. simple integrations.

In this review we will make sure that you are up to speed on all of these basic skills, and we will review trigonometry while we are at it. To get a PDF of this document, click here. The current version (Jan. 19, 2007) has been proof-read and corrected by Donn Henriksen.

There is no substitute for a full year of calculus instruction, and so math 221 is an absolute prerequisite for physics 201, and math 222 is an absolute prerequisite for physics 202. There is no back door.

A smooth function $y = f (x)$ of a variable $x$ is differentiable at $x$ if its derivative

f^{'} (x) = \frac{d f (x)}{d x} = {lim}_{d x \to 0} \frac{f (x + d x) - f (x)}{d x}

exists (is a number) at $x$ . The limit-taking process is very simple; we expand the numerator in ascending powers of $d x$ , perform the division, and take the limit by setting $d x = 0$ afterwards (in a nutshell).

The geometrical interpretation of the derivative of $f (x)$ at $x_{0}$ is that $f^{'} (x_{0})$ is the slope of the line tangent to $f (x)$ at $x_{0}$ ;

Therefore a useful formula for translating calculus to geometry is

tan θ = (\frac{d f (x)}{d x})_{x = x_{0}}

for the tangent to the curve at $x_{0}$ .

The limit taking process is most easily handled for polynomial functions such as

f (x) = \sum_{n = 0}^{N} a_{n} x^{n}

by use of the following simple rule;

\frac{d}{d x} (f (x) + g (x)) = \frac{d f (x)}{d x} + \frac{d g (x)}{d x}

which is easy to prove from the deﬁnition above. This says that the derivative of a (ﬁnite) sum is the sum of the derivatives of the summands.

Example. Let $f (x) = x^{3}$ , the steps taken in computing the derivative are;
Step 1. Write out the fraction

\frac{f (x + d x) - f (x)}{d x} = \frac{{(x + d x)}^{3} - x^{3}}{d x}

= \frac{(x^{3} + 3 x^{2} d x + 3 x {(d x)}^{2} + {(d x)}^{3}) - x^{3}}{d x}

Step 2. Perform the division;

\frac{(x^{3} + 3 x^{2} d x + 3 x {(d x)}^{2} + {(d x)}^{3}) - x^{3}}{d x} = \frac{3 x^{2} d x + 3 x {(d x)}^{2} + {(d x)}^{3}}{d x}

= 3 x^{2} + 3 x d x + {(d x)}^{2}

Step 3. Perform the limit (set $d x = 0$ );

\frac{d x^{3}}{d x} = {lim}_{d x \to 0} (3 x^{2} + 3 x d x + {(d x)}^{2}) = 3 x^{2}

Example. Let $f (x) = a + b x^{2}$ , in which $a$ and $b$ are constants.
Step 1. Write out the fraction

\frac{f (x + d x) - f (x)}{d x} = \frac{(a + b {(x + d x)}^{2}) - (a + b x^{2})}{d x}

= \frac{(a + b (x^{2} + 2 x d x + {(d x)}^{2})) - (a + b x^{2})}{d x}

Step 2. Perform the division;

\frac{(a + b (x^{2} + 2 x d x + {(d x)}^{2})) - (a + b x^{2})}{d x} = \frac{2 b x d x + b {(d x)}^{2}}{d x} = 2 b x + b d x

Step 3. Perform the limit (set $d x = 0$ );

\frac{d}{d x} (a + b x^{2}) = {lim}_{d x \to 0} (2 b x + b d x) = 2 b x

which we can see is the sum of the derivatives of the two terms $a$ and $b x^{2}$ , the derivative of a constant being zero.

1.1 Problems

1 Compute the derivative of

f (t) = x_{0} + v_{0} t + \frac{1}{2} a t^{2}

with respect to $t$ . $x_{0}$ , $v_{0}$ and $a$ are constants. In other words

\frac{d f (t)}{d t} = {lim}_{d t \to 0} \frac{f (t + d t) - f (t)}{d t}

2 Compute the derivative of

f (t) = a {(t - b)}^{4}

with respect to $t$ . $a$ and $b$ are constants.

3 Compute the second derivative of

f (t) = x_{0} + v_{0} t + \frac{1}{3} a t^{3}

with respect to $t$ . This means ﬁrst compute

v (t) = \frac{d f (t)}{d t}

and then compute

\frac{d^{2} f (t)}{d t^{2}} = f^{''} (t) = \frac{d v (t)}{d t}

4 The following problem is very useful in the study of one dimensional motion at constant acceleration. The average velocity of an object over the time interval from $t - \frac{Δ t}{2}$ to $t + \frac{Δ t}{2}$ is

\bar{v} = \frac{x (t + \frac{Δ t}{2}) - x (t - \frac{Δ t}{2})}{Δ t}

with no limit being taken, $Δ t$ can be of any size.
Show that if the average velocity equals the instantaneous velocity at the interval midpoint, namely that if

v (t) = \frac{d x (t)}{d t} = \bar{v} = \frac{x (t + \frac{Δ t}{2}) - x (t - \frac{Δ t}{2})}{Δ t}

then the acceleration is constant. The acceleration is

a (t) = \frac{d^{2} x (t)}{d t^{2}}

2 Derivative rules and formulas; products

Derivatives of a product $f (x) g (x)$ can be easily computed from the deﬁnition,

\frac{d}{d x} (f (x) g (x)) = {lim}_{d x \to 0} \frac{f (x + d x) g (x + d x) - f (x) g (x)}{d x}

by replacing $f (x + d x)$ with $f (x + d x) - f (x) + f (x)$ and rearranging

\frac{d}{d x} (f (x) g (x)) = {lim}_{d x \to 0} \frac{(f (x + d x) - f (x) + f (x)) g (x + d x) - f (x) g (x)}{d x}

= {lim}_{d x \to 0} \frac{(f (x + d x) - f (x)) g (x + d x) + f (x) (g (x + d x) - g (x))}{d x}

= {lim}_{d x \to 0} (\frac{f (x + d x) - f (x)}{d x} g (x + d x)) + f (x) {lim}_{d x \to 0} (\frac{g (x + d x) - g (x)}{d x})

If $f (x), f^{'} (x), g (x)$ and $g^{'} (x)$ all exist, then

= \frac{d f (x)}{d x} g (x) + f (x) \frac{d g (x)}{d x}

We state this as being the product rule for derivatives.

\frac{d}{d x} (f (x) g (x)) = \frac{d f (x)}{d x} g (x) + f (x) \frac{d g (x)}{d x}

2.1 The binomial theorem

This theorem is of great antiquity, and is extremely useful for both algebraic and calculus applications. It says that

{(a + b)}^{N} = \sum_{m = 0}^{N} (\binom{N}{m}) a^{m} b^{N - m}

where the number

(\binom{N}{m}) = \frac{N!}{m! (N - m)!}

is a binomial coefficient, and the factorial of an integer $N$ is

N! = N \cdot (N - 1) \cdot (N - 2) . . . 2 \cdot 1, 1! = 1, 0! = 1

(the last relation is a deﬁnition). For example

{(a + b)}^{2} = a^{2} + 2 a b + b^{2}

{(a + b)}^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

{(a + b)}^{4} = a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + b^{3}

2.2 Problems

5 Use the product rule to show that

\frac{d}{d x} (f (x))^{2} = 2 f (x) f^{'} (x) = 2 f (x) \frac{d f (x)}{d x}

and that in general

\frac{d}{d x} (f (x))^{n} = n (f (x))^{n - 1} f^{'} (x)

2.3 A power tool, series expansion

This last calculation was a little on the tricky side, but there exists a powerful tool for performing most of the operations of calculus in a simple way, the series expansion.

We suppose that the function $f (x)$ exists at the point $x_{0}$ , and for that matter that it exists near $x_{0}$ . Let $x - x_{0}$ be small, so that $x$ is close to $x_{0}$ . The idea of the series expansion is that in the neighborhood of $x_{0}$ , we could replace $f (x)$ with a polynomial

P_{f} (x) = \sum_{n = 0}^{N} \frac{a_{n}}{n!} {(x - x_{0})}^{n}

in which $n! = n \cdot (n - 1) \cdot (n - 2) . . . 2 \cdot 1$ is our factorial of the integer $n$ .

The number of terms $N$ that we need to calculate to get $P_{f}$ depends on what we want to do with it, and is based on the following concept: the function $f (x)$ and polynomial $P_{f} (x)$ agree at $x_{0}$ , and have the same derivative at $x_{0}$ , and the same second derivative, and so on up to the $N^{t h}$ derivative. We would call $P_{f} (x)$ an $N^{t h}$ order series expansion of $f (x)$ about the point $x_{0}$ .
Step 1. Both $f (x)$ and $P_{f} (x)$ agree at $x_{0}$ ;

f (x_{0}) = P_{f} (x_{0}) = a_{0} + a_{1} (x_{0} - x_{0}) + \frac{a_{2}}{2!} {(x_{0} - x_{0})}^{2} + . . . + \frac{a_{N}}{N!} {(x_{0} - x_{0})}^{N} = a_{0}

requires that $a_{0} = f (x_{0})$ .
Step 2. Both $f^{'} (x)$ and $\frac{d}{d x} P_{f} (x)$ agree at $x_{0}$ ;

f^{'} (x_{0}) = 0 + a_{1} + 2 \frac{a_{2}}{2!} (x_{0} - x_{0}) + 3 \frac{a_{3}}{3!} {(x_{0} - x_{0})}^{2} + . . . + N \frac{a_{N}}{N!} {(x_{0} - x_{0})}^{N - 1} = a_{1}

requires that $a_{1} = f^{'} (x_{0}) = {\frac{d f}{d x}}_{x = x_{0}}$ .
Step 3. Both $f^{''} (x)$ and $\frac{d}{d x} P_{f} (x)$ agree at $x_{0}$ ;

f^{''} (x_{0}) = 0 + 0 + 2 \frac{a_{2}}{2!} + 3 \cdot 2 \frac{a_{3}}{3!} (x_{0} - x_{0}) + . . . + N \cdot (N - 1) \frac{a_{N}}{N!} {(x_{0} - x_{0})}^{N - 2} = a_{2}

requires that $a_{2} = f^{''} (x_{0}) = {\frac{d^{2} f}{d x^{2}}}_{x = x_{0}}$ .

For ninety percent of all of the calculus applications in our physics text, this is enough; the polynomial $P_{f} (x)$ that agrees with $f (x)$ up to two derivatives in the neighborhood of $x_{0}$ is

P_{f} (x) = f (x_{0}) + f^{'} (x_{0}) (x - x_{0}) + \frac{1}{2} f^{''} (x_{0}) {(x - x_{0})}^{2} + . . .

This is called the Euler-Maclaurin or Taylor series for $f (x)$ near $x_{0}$ , and it may be substituted in place of $f (x)$ in the neighborhood of $x_{0}$ .

What do we use it for? For starters it can be used to get formulas for derivatives of products and quotients. In most applications you only need to keep one or two terms in a Taylor series. For example, let $x = t + d t$ and $x_{0} = t$ , then

P_{f} (t + d t) = f (t) + f^{'} (t) d t + \frac{1}{2} f^{''} (t) {(d t)}^{2} + . . .

and you can replace any occurrence of $f (t + d t)$ in a formula that involves taking the limit $d t \to 0$ with this expression.

Example

\frac{d}{d t} (f (t) g (t)) = {lim}_{d t \to 0} \frac{(f (t + d t) g (t + d t) - f (t) g (t))}{d t}

= {lim}_{d t \to 0} \frac{(f (t) + f^{'} (t) d t + . . .) (g (t) + g^{'} (t) d t + . . .) - f (t) g (t)}{d t}

= {lim}_{d t \to 0} \frac{g (t) f^{'} (t) d t + g^{'} (t) f (t) d t + . . .}{d t} = g (t) f^{'} (t) + g^{'} (t) f (t)

and we are done quickly and cleanly, all of the terms in $(. . .)$ contain at least two factors of $d t$ , and so in the limit become zero.

Example; l’Hospitals rule is a formula for computing the limit of the ratio of two functions that both vanish at $x_{0}$ ,

{lim}_{x \to x_{0}} f (x) = 0 = {lim}_{x \to x_{0}} g (x)

The limit of the ratio is then

{lim}_{x \to x_{0}} \frac{f (x)}{g (x)} = {lim}_{x \to x_{0}} \frac{f (x_{0}) + f^{'} (x_{0}) (x - x_{0}) + \frac{1}{2} f^{''} (x_{0}) {(x - x_{0})}^{2} + . . .}{g (x_{0}) + g^{'} (x_{0}) (x - x_{0}) + \frac{1}{2} g^{''} (x_{0}) {(x - x_{0})}^{2} + . . .}

but both $f (x_{0}) = 0$ and $g (x_{0}) = 0$ , so

{lim}_{x \to x_{0}} \frac{f (x)}{g (x)} = {lim}_{x \to x_{0}} \frac{f^{'} (x_{0}) (x - x_{0}) + \frac{1}{2} f^{''} (x_{0}) {(x - x_{0})}^{2} + . . .}{g^{'} (x_{0}) (x - x_{0}) + \frac{1}{2} g^{''} (x_{0}) {(x - x_{0})}^{2} + . . .}

divide out the factor $(x - x_{0})$ from numerator and denominator:

= {lim}_{x \to x_{0}} \frac{f^{'} (x_{0}) + \frac{1}{2} f^{''} (x_{0}) (x - x_{0}) + . . .}{g^{'} (x_{0}) + \frac{1}{2} g^{''} (x_{0}) (x - x_{0}) + . . .}

and in the limit $x \to x_{0}$ , $(x - x_{0}) \to 0$ ,

{lim}_{x \to x_{0}} \frac{f^{'} (x_{0}) + \frac{1}{2} f^{''} (x_{0}) (x - x_{0}) + . . .}{g^{'} (x_{0}) + \frac{1}{2} g^{''} (x_{0}) (x - x_{0}) + . . .} = \frac{f^{'} (x_{0})}{g^{'} (x_{0})}

We restate this as l’Hospitals rule;

{lim}_{x \to x_{0}} \frac{f (x)}{g (x)} = \frac{f^{'} (x_{0})}{g^{'} (x_{0})}

provided $g^{'} (x_{0})$ is non-zero. If $g^{'} (x_{0})$ and $f^{'} (x_{0})$ are in fact both zero, we simply repeat the process noting that in

{lim}_{x \to x_{0}} \frac{f^{'} (x_{0}) + \frac{1}{2} f^{''} (x_{0}) (x - x_{0}) + . . .}{g^{'} (x_{0}) + \frac{1}{2} g^{''} (x_{0}) (x - x_{0}) + . . .}

the ﬁrst term in both numerator and denominator are zero and we can divide out another factor of $(x - x_{0})$ ;

= {lim}_{x \to x_{0}} \frac{0 + \frac{1}{2} f^{''} (x_{0}) + . . .}{0 + \frac{1}{2} g^{''} (x_{0}) + . . .} = \frac{f^{''} (x_{0})}{g^{''} (x_{0})}

3 Non-polynomial functions

The most complicated derivatives that you will need to perform are of functions such as

f (x) = \frac{1}{x^{n}}, f (x) = \sqrt[n]{x}, f (x) = sin a x

which are not polynomials. These can be very simply differentiated by using the product rule alone, resulting in differentiation rules for radicals and quotients.

3.1 Rational functions

Consider a function that is the ratio of two functions, both of which you can differentiate;

h (x) = \frac{f (x)}{g (x)}

To compute the derivative of $h (x)$ we take these algebraic steps, ﬁrst

h (x) g (x) = f (x)

now apply the product rule

\frac{d}{d x} (h (x) g (x)) = \frac{d}{d x} f (x) = f^{'} (x), h^{'} (x) g (x) + h (x) g^{'} (x) = f^{'} (x)

and rearrange

h^{'} (x) = \frac{f^{'} (x) - h (x) g^{'} (x)}{g (x)} = \frac{f^{'} (x) - \frac{f (x)}{g (x)} g^{'} (x)}{g (x)} = \frac{f^{'} (x) g (x) - f (x) g^{'} (x)}{g^{2} (x)}

which we will call the quotient rule.

3.2 Radicals

Consider the radical

f (x) = \sqrt[n]{x}

To compute its derivative, ﬁrst raise both sides to the $n^{t h}$ power

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

Now differentiate and apply the product rule repeatedly to the left side

\frac{d}{d x} f^{n} (x) = \frac{d}{d x} x = 1, n f^{n - 1} (x) f^{'} (x) = 1

solve for $f^{'} (x)$ ;

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

We have shown that

\frac{d}{d x} \sqrt[n]{x} = \frac{d}{d x} x^{\frac{1}{n}} = \frac{1}{n} x^{\frac{1}{n} - 1}

and therefore for any power $a$ , integral, rational or otherwise

\frac{d}{d x} x^{a} = a x^{a - 1}

which we call the power rule for differentiation.

Example Find a series expansion for $f (x) = \sqrt{x}$ valid near $x_{0} = 4$ .
The ﬁrst step is to compute a few derivatives, using the power rule with $a = \frac{1}{2}$ ,

\frac{d}{d x} \sqrt{x} = \frac{1}{2} \frac{1}{\sqrt{x}}, \frac{d^{2}}{d x^{2}} \sqrt{x} = \frac{d}{d x} \frac{1}{2} \frac{1}{\sqrt{x}} = - \frac{1}{2^{2}} \frac{1}{\sqrt{x^{3}}}

and so inserting this all into Eq. 5 we ﬁnd that

\sqrt{4 + d x} = \sqrt{4} + \frac{1}{2 \sqrt{4}} d x + \frac{1}{2} (- \frac{1}{2^{2}} \frac{1}{\sqrt{4^{3}}}) {(d x)}^{2} + . . . = 2 + \frac{d x}{4} - \frac{{(d x)}^{2}}{64} + . . .

This should be written using $x = 4 + d x$ , as

\sqrt{x} = 2 + \frac{(x - 4)}{4} - \frac{{(x - 4)}^{2}}{64} + . . .

and in any formula involving $\sqrt{x}$ that will be used for $x$ near $4$ , this is a valid replacement.
In particular, this can be used to calculate square roots of numbers close to $4$ , such as $5$ for which

\sqrt{5} \approx 2 + \frac{1}{4} - \frac{1}{64} + . . . = 2.234375 . . .

which gives quite good accuracy (we are off in the third decimal place) with only these three terms in the series.

3.3 Problems

6 Compute

\frac{d}{d x} \sqrt{x^{2} + 1}

7 Compute

\frac{d}{d x} \sqrt[3]{x^{2} + x}

8 Compute

\frac{d}{d x} (\frac{1}{x^{3} + x})

9 Compute

\frac{d}{d x} (\frac{1 - x^{2}}{x^{3} + x})

10 Find a series expansion for

f (x) = \frac{1}{x^{2}}

valid near $x = 1$ that agrees with $f (x)$ up through the third derivative at $x = 1$ .

11 Find a series expansion for

f (x) = \frac{1}{{(1 - x)}^{2}}

valid near $x = 0$ that agrees with $f (x)$ up through the third derivative at $x = 0$ .

12 There is a certain function with the truly unique property that at any point $x$ , all of its derivatives are the same;

f (x) = f^{'} (x) = f^{''} (x) = f^{'''} (x) = . . .

If we deﬁne $f (0) = 1$ , ﬁnd the Taylor series expansion for $x$ near zero for this function.

13 Show that

\sum_{n = 0}^{N} x^{n} = 1 + x + x^{2} + . . . + x^{N} = \frac{1 - x^{N + 1}}{1 - x}

This can be done by purely elementary means.
Use l’Hopital’s rule to show that

\frac{d}{d x} \sum_{n = 0}^{N} x^{n} |_{x = 1} = \sum_{n = 0}^{N} n = \frac{N (N + 1)}{2}

14 Use l’Hopital’s rule to compute

{lim}_{x \to 1} x^{N} -