From Euclid to Spigotry
A Quick Review of Calculus
Last modified: September 9, 2004
Lecture topic 1: Definition, linearity and chain rule:
State two equivalent definitions for the derivative f'(x).  From the definition, show that 
a. if h(x) = a f(x) + b g(x) then h'(x) = a f'(x) + b g'(x)  (linearity)
b. if f(x) = x2, then f'(x) = 2x.
c. if h(x) = f(g(x)) then h'(x) = f'(g(x))g'(x) (chain rule)
d. if h(x) = x4 then h'(x) = 4 x3.

Notes:
The definition is f'(x) =  It is sometimes convenient, just to make the
algebra work out nicely, to use the equivalent definition f'(x) =  
a. Linearity is easy. 
b. Use the second form of the definition and cancel a common factor of y - x.
c. For the chain rule write
hp'(x) =and use the fact (not proved) that the limit of a product equals the product of limits.
d. Notice that h(x) = f(f(x)) where f(x) = x2. Use the chain rule. Don't expand (y-x)4, since that approach will be used next.
Lecture topic 2: The integer power rule and the product rule
a. Using the factorization 
show that if f(x) = x3 then f'(x) = 3 x2.
b. By writing 4 f(x)g(x) =  
and differentiating both sides (using the chain rule on the right-hand side)
show that if h(x) = f(x)g(x) then h'(x) = f'(x)g(x) + f(x)g'(x).
c. Using this "product rule" prove by induction that
if f(x) = xn for positive integer n, then f'(x) = n xn-1.  Hint: xn+1 is the product of x with xn.

Lecture topic 3: Negative exponents and the quotient rule
a. By writing g(x) = and differentiating f(x)g(x) = 1, show that g'(x)= -f'(x)/(f(x))2.
By setting f(x) = x, show that if g(x) = x-1 then g'(x) = - x-2
b. Using the chain rule with f(x) = xm , g(x) =  , and h(x) = f(g(x)), show, using the result
 g'(x) = - x-2 from above, that for any positive integer m, if h(x) = x-m then h'(x) = -m x-m-1
c. Prove the quotient rule: if h(x) = f(x)/g(x)  then h'(x) = (g(x)f'(x)-f(x)g'(x))/(g(x)2)

Lecture topic 4: Differentiating trig functions
a. Using the definition of the derivative, and applying the addition formula 
sin(x+e) = sin x cos e + cos x sin e,
show that if f(x) = sin x then f'(x) = cos x.
b. Using the definition of the derivative, and applying the addition formula 
cos(x+e) = cos x cos e - sin x sin e,
show that if f(x) = cos x then f'(x) = -sin x.
c. By using the minus-1 power rule, show that if f(x) = sec x, f'(x) = sec x tan x.
d. By writing tan x = sin x sec x and using the product rule, show that 
if f(x) = tan x then f'(x) = sec2 x.
Lecture topic 5: The inverse-function rule
Suppose that f is a function that "undoes" g, so that f(g(x)) = x.
Example: g(x) =  while f(x) = x2.
a. Use the chain rule to show that g'(x) = 1/ f'(g(x)).

b. Show that if g(x) =  then g'(x) = 
c. Using the chain rule with f(x) = xm and g(x) =  show that
if h(x) =  then h'(x) = . This completes the proof of the rule for differentiating a
power of x when x is a rational number, positive or negative. We will need this in order
to establish the general case of the binomial theorem, which Newton formulated but did not prove.
Lecture topic 6: Inverse trig functions and logarithms
 
a. Show that if g(x) = arcsin x (so that sin(g(x)) = x)

then g'(x) = 
b. Show that if g(x) = arctan x (so that tan(g(x)) = x)

then g'(x) = 
c. Define the exponential function f(x) = ex by the properties that f'(x) = f(x) and f(0) = 1.
Show that for positive x, if g(x) = log x (so that eg(x) = x) then g'(x) = 

Lecture topic 7: Integrals as areas
Define  , where b > a, as the area under the graph of f(x) from x =a to x =b.
To keep things really simple, assume that f(x) is everywhere positive.

a. By drawing pictures, argue that +=
b. The "integral form of the mean-value theorem," which we shall need from time to time, states that if f(x) is a continuous function (one whose graph can be drawn without lifting the pencil off the paper) on the interval (a,b), then there is at least one X satisfying a < X < b 
such that  = f(X)(b-a). Present an informal proof of this result based just on the concept of integral as area. Sketch an example to show how the result may not be true if f(x) is discontinuous at just a single point in (a,b).
Lecture topic 8: The derivative of an integral: the integral of a derivative
Define F(x) =  (assume that x > 0)
a. Using the first definition of the derivative from topic 1 and the mean-value theorem from topic 7b, argue that F'(x) = f(x). This is the famous "fundamental theorem of calculus."
b. Suppose that 0 < a < b. Using the result of topic 7a, argue that
 = F(b) - F(a).      This is another version of the "fundamental theorem of calculus."
c. Let f(x) = F'(x) and g(x) = G'(x). You already know, from topic 1, that if p and q are constants and H(x) = p F(x) + q G(x), then H'(x) = p F'(x) + q G'(x) = p f(x) + q g(x). Prove that
 + q
(Surprisingly, this is hard to prove directly from the definition of integral as area.)
d. Given that if F(x) = xn+1 then F'(x) = (n+1) xn, show that  = (bn+1- an+1)
and that for non-negative n,  = 
This is the integration formula that you need in order to understand how Newton and Gregory
approximated p by means of infinite series. Although the important case is the one where n is a non-negative integer, our proof is valid for any rational value of n, positive or negative, except
n = -1. Show that topic 6c takes care of that case by the formula
 = log b - log a

Lecture topic 9: The mean-value theorem
Starting from the integral form of the mean-value theorem from topic 7b, prove the more familiar form of the mean-value theorem: if F(x) is differentiable everywhere on (a,b) and is
continuous at a and b, then there is at least one X satisfying a < X < b such that

F'(x) = 
Give an example showing that F(x) really must be differentiable everywhere on (a,b)
You might take F(x) =  with a = -1, b = 1.
Lecture topic 10: Length of a curve as an integral
Since the best-known definition of p is in terms of the circumference of a circle, it is useful to have
a formula for the length of a curve.
Let L(x) denote the length of the graph of y = f(x) between 0 and x (assume x > 0). Then the derivative of this function is
L'(x) = .
Now L(x+h) - L(x) is just the length of the short portion of the graph between x and x+h.
This argument is a bit trickier than the one for areas.
Assume that the derivative of f(x) is always 0 or positive in the interval (x, x+h). (If it is negative, the same argument will work, but some signs need to be changed.)

Let m denote the smallest value of the derivative f' and M its largest value on [x, x+h].
By sketching a graph of f, argue that its length is at least as great as the length of a line of slope m and at least as small as the length of a line of slope M, and that therefore there must be at least one point X in (x, x+h) for which the length of the graph of f is equal to the length of a straight line whose slope is f'(X). By the Pythagorean theorem, this length is .

Now you can conclude that L'(x) = 
and so, by the "fundamental theorem," the length of the graph of y = f(x) between 0 and a is
L(a) =