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Lecture Notes on Differentiation
A tangent line to a function at a point is the line that best approximates the function at that point better than any other line.
The slope of the function at a given point is the slope of the tangent line to the function at that point.
The derivative of f at x = a is the slope, m, of the function f at the point x = a (if m exists), denoted by f ′(a) = m. All other notations:
y′, dydx , dfdx , (^) dxd f (x), Dxy, Dxf (x).
The function f (x) is differentiable at a point x 0 if f ′(x 0 ) exists. If a function is differentiable at all points in its domain (i.e. f ′(x) is defined for all x in the domain), then we consider f ′(x) as a function and call it the derivative of f (x).
The derivative of f that we have been talking about is called the first derivative. Now, we define the second derivative of a function to be the derivative of f ′, denoted by f ′′(x)
or d
(^2) f dx^2 (=^
d dx
( (^) d dx f^ )
Example 1: Given f (x) = c where c is a constant. Then f ′(x) = 0 because the slope of the function at each point is zero.
Example 2: If f (x) = 2 − 3 x , then the derivative f ′(x) = 2 because the slope of the function at each point is 2.
Example 3: Given f (x) = |x|. We have
f ′(x) =
− 1 if x < 0 1 if x > 0
However, f ′(0) is not defined because there is no unique tangent line to f (x) at x = 0.
The following is a table of derivatives of some basic functions:
f (x) f ′(x) c 0 mx + c m xa^ axa−^1 ex^ ex ln x (^) x^1
Rules of Differentiation:
- (f ± g)′^ = f ′^ ± g′
- (c · f )′^ = cf ′
- (Product Rule) (f · g)′^ = f ′g + f g′
- (Quotient Rule)
f g
f ′g − f g′ g^2
(where g(x) 6 = 0)
- (Chain Rule) (f ◦ g)′^ = (f (g(x)))′^ = f ′(g(x)) · g′(x)
The equation of the tangent line to the function at point x = x 0 is:
y − f (x 0 ) = f ′(x 0 )(x − x 0 )
Theorem (The Extreme-Value Theorem for Continuous Functions) If f is continuous at every point of a closed interval I, then f assumes both an absolute maximum value value M and an absolute minimum value m somewhere in I.
Definition A point in the domain of a function f at which f ′^ = 0 or f ′^ does not exist is a critical point of f.
Theorem Extreme values (local or global) occur only at critical points and endpoints.
Examples:
- Find absolute maximum and minimum values of f (x) = 4 − x^2 on the interval [− 3 , 1].
- Find absolute maximum and minimum values of f (x) = x^2 /^3 on the interval [− 1 , 8].
- Find absolute maximum and minimum values of f (x) = x^1 /^3 on the interval [− 1 , 1].
Theorem (The Mean Value Theorem) Suppose the f (x) is continuous on a closed interval [a, b] and differentiable on the interval’s interior (a, b). Then there is at least one point c in (a, b) at which
f (b) − f (a) b − a
= f ′(c).
Below is a picture illustrating when a function is concave up or concave down. Notice the tangent lines and their slopes. A point of inflection is also labeled on the picture.
Note: To find the inflection points, we look at the second derivative. Find all the points such that f ′′^ is zero or undefined at those points. Then use the Key Number Method to test the sign changes of f ′′^ at those points.
Examples:
- f (x) = x^3 − 12 x − 5.
- f (x) = x^4 − 4 x^3 + 10.
Examples from Economics
Suppose that r(x) = the revenue from selling x items c(x) = the cost of producing the x items p(x) = r(x) − c(x) = the profit from producing and selling x items.
The marginal revenue, marginal cost, and marginal profit when producing and selling x items are the derivatives dr dx
= marginal revenue, dc dx
= marginal cost, dp dx
= marginal profit.
Let’s consider the relationship of p to these derivatives. If r(x) and c(x) are differentiable for all x > 0, and if p(x) = r(x) − c(x) has a maximum value, it occurs at a production level at which p′(x) = 0. Since p′(x) = r′(x)−c′(x), p′(x) = 0 implies that r′(x) − c′(x) = 0 or r′(x) = c′(x).
Therefore,
At a production level yielding maximum profit, marginal revenue equals marginal cost.
Figure. The graph of a typical cost function starts concave down and later turns concave up. It crosses the revenue curve at the break-even point B. To the left of B, the company operates at a loss. To the right, the company operates at a profit, with the maximum profit occurring where c′(x) = r′(x). Farther to the right, cost exceeds revenue (perhaps because of a combination of rising labor and materials costs and market saturation) and production levels become unprofitable again.
Lecture Notes on Integration
Mean Value Theorem Suppose f (x) is continuous on [a, b] and differentiable on (a, b). Then there exists a point c in (a, b) at which
f (b) − f (a) b − a
= f ′(c). (1)
Corollary 1 If f ′(x) = 0 at each point of an interval I, then f (x) = C for all x in I, where C is a constant.
Corollary 2 If f ′(x) = g′(x) at each point of an interval I, then there exists a constant C such that f (x) = g(x) + C for all x in I.
A function, F (x), is an antiderivative of a function f (x) if F ′(x) = f (x) for all x in the domain of f.
Example: The function F (x) = x^2 is an antiderivative of f (x) = 2x. The function G(x) = x^2 + 4 is also an antiderivative of f (x) = 2x.
The set of all antiderivative of f is the indefinite integral of f with respect to x, denoted by (^) ∫
f (x)dx
The symbol
is an integral sign. The function f (x) is the integrand of the integral, and x is the variable of integration.
To verify
xexdx = xex^ − ex^ + C, we take the derivative of the right hand side. d dx
xex^ − ex^ + C = ex^ + xex^ − ex^ = xex. Thus, the integral statement is correct.
∫ Integral formulas xndx =
xn+ n + 1
dx = x + C ∫ exdx = ex^ + C ∫ 1 x
dx = ln |x| + C
Rules for indefinite integrals:
kf (x)dx = k
f (x)dx
−f (x)dx = −
f (x)dx
[f (x) ± g(x)]dx =
f (x)dx ±
g(x)dx
Method of Substitution: ∫ f (g(x))g′(x)dx =
f (u)du where u = g(x) and du = g′(x)dx.
Example 1: Find
(x^3 + 2)^53 x^2 dx.
Let u = x^3 + 2, du = 3x^2 dx. Then ∫ (x^3 + 2)^53 x^2 dx =
u^5 du
u^6 6
+ C
(x^3 + 2)^6 6
+ C
Example 2: Find
x^2 + 1 · 2 xdx.
Let u = x^2 + 1, du = 2xdx. Then ∫ (^) √ x^2 + 1 · 2 xdx =
udu
u (^32) 3 2
+ C
2 u (^32)
3
+ C
2(x^2 + 1) (^32)
3
+ C
Definition: (Definite Integral)
∫ (^) b
a
f (x)dx = (signed or net) area between the curve and x-axis from a to b.
The number a is called the lower limit and the number b is called the upper limit.
Properties for definite integrals:
∫ (^) b
a
f (x)dx = −
∫ (^) a
b
f (x)dx
∫ (^) a
a
kf (x)dx = 0
∫ (^) b
a
kf (x)dx = k
∫ (^) b
a
f (x)dx
∫ (^) b
a
−f (x)dx = −
∫ (^) b
a
f (x)dx
∫ (^) b
a
[f (x) ± g(x)]dx =
∫ (^) b
a
f (x)dx ±
∫ (^) b
a
g(x)dx
∫ (^) b
a
f (x)dx +
∫ (^) c
b
f (x)dx =
∫ (^) c
a
f (x)dx
- min f · (b − a) ≤
∫ (^) b
a
f (x)dx ≤ max f · (b − a)
- If f (x) ≥ g(x) on [a, b], then
∫ (^) b
a
f (x)dx ≥
∫ (^) b
a
g(x)dx
We can approximate the area under the curve using rectangles.
Left Endpoint Rule: using rectangles with left top corner on the curve Right Endpoint Rule: using rectangles with right top corner on the curve Midpoint Rule: using rectangles with top midpoint on the curve
Ln = sum of area of n rectangles using left endpoint rule. Rn = sum of area of n rectangles using right endpoint rule. Mn = sum of area of n rectangles using midpoint rule.
The Fundamental Theorem of Calculus (Part 1) Let f be a continuous function on [a, b]. Let F be the function
F (x) =
∫ (^) x
a
f (t)dt.
Then, F (x) is continuous on [a, b], differentiable on (a, b) and
dF dx
d dx
∫ (^) x
a
f (t)dt = f (x).
(Part 2) If f is a continuous function on [a, b] and F is any antiderivative of f on [a,b], then
∫ (^) b
a
f (t)dt = F (b) − F (a) := F (x)
b a
Example 1:
d dx
∫ (^) x
1
t^2 + 3dt =
x^2 + 3
Example 2:
d dx
∫ (^) x^2
3
tetdt = x^2 ex
2 · (2x) = 2x^3 ex
2
Example 3: ∫ (^2)
1
(3x^2 + 2x)dx = x^3 + x^2
2 1
Example 4: ∫ (^2)
0
2 x
x^2 + 1dx
First find an antiderivative of 2x
x^2 + 1. ∫ 2 x
x^2 + 1dx
Let u = x^2 + 1, du = 2xdx.
Then
2 x
x^2 + 1dx =
udu =
2 u (^32)
3
+ C =
2(x^2 + 1) (^32)
3
+ C.
Thus,
0
2 x
x^2 + 1dx =
2(x^2 + 1) (^32)
3
1 0
(^32)
3
