MATH 116-2003
Derivative & Antiderivative review
This review covers (most of) the derivative/antiderivative material from first semester that we will call upon for second semester (doesn't include much on graphs & calculation with log & exponential, exponential growth functions - which also will show up quite a bit). This list is intended to remind you of derivative ideas covered so that you can see where you need to find more practice problems - it does not include enough exercises for relearning techniques if you have trouble.
In these notes, the notation Dxf is used for "the derivative of the function f with respect to the variable x" which is also written or (if the variable is understood) f¢ .
Review exercises
(should be done by 1/17/2003, but will not be collected) In Barnet and Ziegler:
p217 1, 23, 29-32, 81, 82 (Remember to include units in the answers for 81 & 82)
p254 19, 55, 57
p270 89 (add: E. What is the population size when the population is growing fastest?)
p304 11, 54
p309 1-8, 29, 30, 68
p342 15, 17, 39, 41, 43, 73, 87, 88
p350 21, 35
p373 49, 57, 59, 71, 73, 103
p387 5, 9, 17, 23, 31, 41, 43, 55, 75
NOTES
Derivatives
Derivative is a rate of change (In particular, if N(t) is size at time t , then DtN gives growth rate ): because slope also represents rate of change, Dxf gives the slope of the tangent to the graph of y = f(x)
I. Derivative Rules
Rules for finding the derivative from a formula:(u and v represent any functions of the variable x , or any expressions involving the variable x)
A.) The big five (written in the most general form)[there are more - but these are by far the most important]:
1.) Power rule: Dxun = nun-1Dxu
2.) Exponential rule: Dxeu = eu Dxu (For base a : Dxau = au Dxu ln a)
3.) Logarithm rule: Dxln u = Dxu (For base a : Dxlogau = Dxu logae)
*4.) Sine rule: Dx sin u = cos u Dxu [To be covered second semester]
*5.) Cosine rule: Dx cos u = -sin u Dxu [To be covered second semetster]
B.) Reduction rules:
1.) Constant coefficient rule: Dx(ku) = k Dx(u) for any constant k
2.) addition/subtraction rule: Dx(u + v) = Dxu + Dxv and Dx(u - v) = Dxu - Dxv
3.) Product rule: Dxuv = v Dxu + u Dxv [or : h0Dhi + hiDh0]
4.) Quotient rule Dx=
5.) Chain rule: Dxf(u) = Duf(u) Dxu
Examples [everything uses the chain rule]:
a.) Dx(8x3e5x) = 8(e5x Dx(x3) + x3Dxe5x) = 8(e5x(3x2) + x3(e5x(5)) ) = [Uses constant coefficient, product, power, exponential rules]
b.) Dte = e Dt(5t2) = e = [Uses exponential, constant coefficient, power rules)
c.) Dx(3x2 + 1)3 = 3(3x2 + 1)2Dx(3x2 + 1) = 3(3x2 + 1)26x = [Uses power, constant coefficient, addition rules]
d.) Dv ln(v3 - 5v) = = =
[Uses logarithm, power, subtraction rules]
e.) Dt 3t sin(5t2- t) = sin(5t2-t) Dt(3t) + 3t (Dt sin(5t2-t)) = sin(5t2-t) (3) + 3t (cos(5t2-t) Dt(5t2-t)) = [uses product, constants coefficient, power, sine, subtraction rules]
Everything uses the chain rule - it's even built in to the big five, in their general forms. Remember this: it is very important in finding antiderivatives
Remember that Dxx = 1, Dx(anything else) ≠ 1 (same for any other variable)
II. Implicit Differentiation:
If A = B, then DxA = DxB (Used with the chain rule to get derivatives, when we can't solve an equation for y - or [usually used with Dt , in this case] for getting rate-of-change relations from relations on quantities )
Example:
If 3x2 + ln y - xy2 = 8, we can get Dxy at any point (x,y) on the graph, by implicit differentiation:
Dx(3x2 + ln y - xy2) = Dx8
3(2x) + Dxy - (y2 + x(2yDxy)) = 0
6x - y2 + Dxy - 2xyDxy = 0
( - 2xy)Dxy = -6x + y2
[Need to know both x and y to get numerical value - important ideas are: 1. taking derivative of both sides of equation; 2. keeping Dxy as a symbol (a new variable) since we don't have a formula for it during the work]
Antiderivatives
A function F(x) is an antiderivative of f(x) if DxF(x) = f(x). The general antiderivative of a function always contains an undetermined constant (represented as C ) so that it represents all antiderivatives at once. To get a particular antiderivative, you need to know the value of the function (antiderivative) at one point.
The antiderivative rules all come from the derivative rules, but using them is not as straightforward because the chain rule kicks in from the beginning. We use to represent "the general antiderivative of f(x) with x as variable" Since "the general antiderivative" of a derivative is the origional function plus or minus any constant, we have . The rules are often written in a shorter form in which u represents any function of x , and du represents u¢dx (that is, Dxu dx) thus
A. The big five:
1.) Power rule (nice case) = + C if n ≠ -1 (covers other negative exponents, though ) (Remember special appearance when n=1 : = + C )
[Short form: ]
2.) Power rule (not nice case : sometimes called special fraction rule)
= = = ln + C
[Short form: ]
3.) Exponential rule = eu + C
[Short form ]
*4.) sine rule: = - cos u + C [New for second semester]
[Short form: ]
*5.) cosine rule: = sin u + C [New for second semester]
[Short form:]
B. Reduction rules:
1.) Constant coefficient rule: = k for any constant k (Constants can filter out through the integral symbol - though variables cannot)
2.) Sum/difference rules = ±
C. The "change of variables" (also called "u-substitution") technique: The formulas are interpreted to mean that, for example any un can be integrated if we have du , but not otherwise. We can use the constant coefficient rule to take care of missing constant coefficients (because the adjustments filter out to the front of the antiderivative), but nothing else can be missing (or added). Every antiderivative computation that we can handle at this stage uses one of the "big five" rules, - or a table rule, as we will see later. The "change of variables" technique is based on the chain rule and is used all the time, just as the chain rule is used in every derivative situation.
Examples:
a.) = with u = x2 (because if u = x2 , then u¢ is Dx(x2) = 2x so du = 2x dx , which is exactly what we have - notice it's OK that the 2x and the dx are not right next to each other in the integral, as long as they are multiplied) thus we use the exponential rule and get
= = e+ C =
b.) is almost with n = 5, but u has to be (3t2-2t) to give us a match, so du = (6t-2)dt = 2 (3t-1)dt and what we have is (3t-1)dt . Since all that is missing is the 2 (a constant), we can fix this:
= = = + C =
[This uses "nice" power rule]
c.) = =
(with u = ex+x, so that u¢ = (ex+1) ) = ln + C =
[Uses the "less nice" power rule. In this case, the absolute value can't be dropped because we don't know whether ex+x is positive or negative]
d.) = = = (with u = 3x2-1, u¢ = 6x) = + C = + C =
[This uses the "nice" power rule, even though we have a fraction, because it fits both conditions: 1. the denominator is a power and 2. the numerator is (except for a constant) the deriviative of the base of the power]
We still have to see what we get from the derivative rules for sine & cosine and from products. The quotient rule does not give any useful antiderivative technique.
Uses/related ideas:
1. Whenever Dxf is positive, f is increasing (values getting larger: graph is rising). Whenever Dxf is negative, f is decreasing (values getting smaller: graph is falling)
There will be a relative (local) maximum for f where Dxf changes from positive to negative, and a relative minimum for f where Dxf changes from negative to positive (as long as f is continuous at the point of change)
Example: If the graph of Dxf looks like this:
Then f is increasing for x < a and for b < x < c , because Dxf is positive in these intervals, and f is decreasing for a < x < b because Dxf is negative in these intervals. There is a relative maximum for f (but we don't know its value) at x = b , and there are local minima (but we don't know the values) at x = a and at x = b.
2.) If we have a formula r(t) for the rate of change of some quantity N(t), then r(t) = N'(t) and N(t) = = - but we need a value of N at some value of t to get the particular antiderivative.
Example: If a bacterial population grows so that the growth rate (by mass) is given by r(t) = 1.2e.01t mg/hr, and the mass after 4 hours is 190 mg , we can get the formula for the size N(t) by taking N(t) = = e.01t + C , and using the known size (and some arithmetic) to get 190 = N(4) = 120e.01(4) + C so that C = 190-121=69 and