11 March 2005

What does f ' (n) mean?

by George E. Hrabovsky, President, MAST

Where We Have Been

Last time we actually calculated an integral. This is one of the prominent tools of calculus. We have answered a lot of the questions posed since, "What is a function?" Now we turn to another prominent tool of calculus, the derivative.

Session 8: What does f ' (n) mean?

For this I will introduce a new source, [1], and gain the following definition,

f ' (x) = Underscript[lim, h0] (f(x + h) - f(x))/h .

Mr. Dawkins informs us that this is the definition of the derivative and that the derivative tells us how far the function of x, f(x), changes as x changes by some extremely small number in the limit of zero, h. We can rewrite this in a language more familiar to use from our integration of last week,

f ' (x) = Underscript[lim, Δ x0] (f(x + Δ x) - f(x))/(Δ x) .

If we examine the numerator on the right hand side,

f(x + Δ x) - f(x)

We can see that this is an interval, as our function goes from some value f(x) to some later value f(x + Δ x).  This interval is the quantity that the function changes as x changes.  So, we can write

Δ y = f(x + Δ x) - f(x) .

Thus, our derivative becomes

f ' (x) = Underscript[lim, Δ x0] (Δ y)/(Δ x) .

We can see that the derivative is the ratio of two intervals as one interval becomes infinitely small.  Put another way, the derivative allows us to write the change in a dependent variable as the independent variable changes.  In this way we can see that the derivative is itself a function.

60. How can we use derivatives?

How can we use derivatives?

We can use our definition to create several rules that we will find useful.

Constant Rule

If,

f(x) = c

where c is a constant, we can apply our definition of the derivative to determine that,

f ' (x) = Underscript[lim, Δ x0] (f(x + Δ x) - f(x))/(Δ x)

              = Underscript[lim, Δ x0] (c - c)/(Δ x)

              = Underscript[lim, Δ x0] 0/(Δ x)

FormBox[RowBox[{             , RowBox[{=,  , RowBox[{Underscript[lim, Δ x0] 0, =, 0.}]}]}], TraditionalForm]

This is understandable since the rate of change of a constant is 0.

Constant Multiple Rule

If,

f(x) = c x

where c is a constant, we can apply our definition of the derivative to determine that,

f ' (x) = Underscript[lim, Δ x0] (f(x + Δ x) - f(x))/(Δ x)

              = Underscript[lim, Δ x0] (c (x + Δ x) - c x)/(Δ x)

              = Underscript[lim, Δ x0] (c x + c Δ x - c x)/(Δ x)

              = Underscript[lim, Δ x0] ( c Δ x )/(Δ x)

              = Underscript[lim, Δ x0] c   = c .

This tells us that the rate of change of a constant multiple is the constant.

Sum Rule

If,

f(x) = u + v

where u and v are dependent on x, we can apply our definition of the derivative to determine that,

f ' (x) = Underscript[lim, Δ x0] (f(x + Δ x) - f(x))/(Δ x)

              = Underscript[lim, Δ x0] ((u + v) (x + Δ x) - (u + v) (x))/(Δ x)

              = Underscript[lim, Δ x0] (u(x + Δ x) + v (x + Δ x) - u(x) - v(x))/(Δ x)

              = Underscript[lim, Δ x0] (u(x + Δ x) - u(x) + v (x + Δ x)    - v(x))/(Δ x)

              = Underscript[l ... 6; x) - u(x))/(Δ x) + Underscript[lim, Δ x0] (v(x + Δ x) - v(x))/(Δ x)

              = u ' (x) + v ' (x) .

This tells us that the derivative of a sum is the sum of the derivatives.

Product Rule

If,

f(x) = u v

where u and v are dependent on x, we can apply our definition of the derivative to determine that,

f ' (x) = Underscript[lim, Δ x0] (f(x + Δ x) - f(x))/(Δ x)

              = Underscript[lim, Δ x0] ((u v) (x + Δ x) - (u v) (x))/(Δ x)

              = Underscript[lim, Δ x0] (u v x + u v Δ x - u v x)/(Δ x)

              = Underscript[l ... ; x) + Underscript[lim, Δ x0] (v (u x +   u Δ x) - v (u x))/(Δ x)

              = uUnderscript[ ... Δ x) + vUnderscript[lim, Δ x0] (u (x +   Δ x) - u (x))/(Δ x)

              = u v ' (x) + v u ' (x) .

Power Rule

If,

f(x) = x^n

where n is some power, we can apply our definition of the derivative to determine that,

f ' (x) = Underscript[lim, Δ x0] (f(x + Δ x) - f(x))/(Δ x)

              = Underscript[lim, Δ x0] ((x + Δ x)^n - (x)^n)/(Δ x) .

We can apply the binomial theorem here,

(a + b)^n = Underoverscript[∑, k = 0, arg3] (n) a^kb^(n - k),                                                     k

where,

(n) = n !/((n - k) ! k !),   k

and,

n ! = 1 · 2 · 3 · ... · n .

So the zero term becomes

b^n

The next term is

n a b^(n - 1)

the next term is

((n - 1) n)/2 a^2 b^(n - 2)

the middle term is

n !/((n - k) ! k !) a^kb^(n - k)

the third to the last term is

((n - 1) n)/2 a^(n - 2) b^2

the second to the last term is

n a^(n - 1) b

and the last term is

a^n .

We can put this together,

(a + b)^n = a^n + n a^(n - 1) b + ((n - 1) n)/2 a^(n - 2) b^2 +... + n !/((n - k) ! k !) a^kb^(n - k) +... + ((n - 1) n)/2 a^2 b^(n - 2) + n a b^(n - 1) + b^n .

Returning to our derivative

f ' (x) = Underscript[lim, Δ x0] ((x + Δ x)^n - (x)^n)/(Δ x)

              = Underscript[l ...  k) +... + ((n - 1) n)/2 x^2 Δ x^(n - 2) + n x Δ x^(n - 1) + Δ x^n - x^n) .

              = Underscript[l ... (n - k) +... +  ((n - 1) n)/2 x^2 Δ x^(n - 2) + n x Δ x^(n - 1) + Δ x^n ) .

              = Underscript[l ... ) +... +  ((n - 1) n)/2 x^2 Δ x^(n - 3) + n x Δ x^(n - 2) + Δ x^(n - 1) ) .

              = n x^(n - 1) .

What does f ' (1) mean?

Just like f(1) means the value of the function f at 1, f ' (1) is the value of the derivative at 1.  For example,

f(x) = x^2

implies

f ' (x) = 2x

by the power rule.  So , if we want to see the behavior of these functions when x = 1, we have

f(1) = 1^2 = 1,

and

FormBox[RowBox[{f ' (1),  , =,  , RowBox[{2 · 1, =, 2.}]}], TraditionalForm]

Book Review: Complete Calculus 1

Paul Dawkins,  "Complete Calculus I," http://tutorial.math.lamar.edu/, 2004.

This electronic book is 348 pages long and is reasonably good, especially for the price. The first 53 pages constitute a short and intense course in algebra and trigonometry.

The chapter on limits tries to motivate the idea of limits with a nine page set of examples that force you to the idea of limits. Limits are then introduced and applied to a number of functions. The idea that the value of a limit can change with its direction of approach is then introduced. This the one-sided limit. This is followed by a list of the fundamental principles of limits. A discussion of how to calculate limits is then followed by a presentation of the properties of infinite limits. This is followed by the notion of continuity culminating in the famous Intermediate Value Theorem. This chapter is wrapped up with the formal definition of a limit. This chapter is just over 50 pages long.

The chapter on derivatives begins with a definition similar to that above. This is followed by a discussion of how to interpret derivatives. There is a presentation of many of the principles of differentiation (like the sum rule and so forth.) These are applied to many different types of functions. The chain rule is presented and then the idea of implicit differentiation is explored. The ideas of related rates and higher-order derivatives are then covered. A simplification scheme called logarithmic differentiation is covered next. We see many standard applications covered in another chapter; the identification of critical points, maximum and minimum values of functions, curve sketching, the mean value theorem, optimization, indeterminate forms, the linear approximation and differentials, and Newton's method.  In all, these two chapter constitute over 120 pages.

The chapters on integration are every bit as extensive.  The book wraps up with what it calls extras that provide proofs to some theorems and some of the deeper background.

Reference

[1] Paul Dawkins, "Complete Calculus I," http://tutorial.math.lamar.edu/, 2004.


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