Calculus Lecture - Ch 7
3.6 Chain
Rule
Why do 3.6 in 4B? Support int by subst.
Purpose – deal with composite functions f(g(x)). Note inner vs outer functions: learn to recog
One way to determine is to evaluate a function for a specific value of the variable; the first calculation is with the innermost function, etc., and the last calculation is with the outermost function.
Deriv is f’(g(x))*g’(x) Or, for clarity, deriv is f’(g)*g’. Or, f’(g)*Dx(g)
Ex: Dxsin(sqrt x) What happens to inner and outer funcs?
Outer is sin(u), where u=x^(1/2)
Intermed step: cos(sqrt x)*Dxsqrt x Note repeat the inner func; ans is a product.
App: find eqn of tang line to this curve at x=p2/36.
y-1/2 = 3sqrt(3) / (2p) (x-p2/36)
Leibniz notation: for functions y=y(u) and u=u(x), then dy/dx = dy/du * du/dx Note canx.
Also note we’ve avoided introducing extra letters.
Ex: Dxtan3(2x). Note special trig notation and extra step in chain: Outer is u3, u=tan 2x
App: check by comparing graphs of func and deriv
Combine with product/quotient rule
Ex: y=(2x-5)^4*(8x^2-5)^(-3) Ans: 8(2x-5)^3*(8x^2-5)^(-4)*(-4x^2+30x-5)
Further examples:
Ex (know your defs!). Suppose
F(x)=f(g(x)), f(2)=1, f(5)=4, f'(2)=3, f'(5)=−1,
g(2)=5,g(5)=7, g'(2)=−1,
g'(5)=0. Find F'(2). Ans: f'(g(2)*g'(2) = f'(5)*3 = 4*3 = 12
Explain result graphically
Ex (~X63, plus...) know defs and be able to read deriv from graph: undef at corner
Ex (Demo CAS): How to define for Maple [note: you can copy and paste this into Maple]:
Task: Find deriv, in factored form, and determine where function has HTL or a VTL:
f:=sqrt(x^2-2*x-15)/(x+4)^3;
diff(f,x);
g:=factor(%);
h:=solve(g=0,x);evalf(%);
solve(denom(g)=0);
plot(f,x=-6..10, y=-.004..0.004);
plot(f,x=-6..-2.5,y=-20..20);
eval(f,x=h[1]);evalf(%);
5.3 Integration by Substitution Hwk: Also do òx3/(x2+2)dx: start with long division.
Simplest case: the argument of the function is linear:
Divide by the coefficient of x
Typical for integrands that are products of two factors:
u-sub for one factor immediately clears out the other factor
Ex: if integrand is sec3(x)tan(x), use sec x, so you get u^2, with du = sec x tan x
More complicated: ![]()
Solve the eqn that defines u (for x or some function of x) and make further substitutions.
Other examples:
![]()
![]()
![]()
Long Div if degree in num ³ degree in denom:
Example
: becomes x - 3x/(x2+3)
Use formula
![]()
Example
: Trick: add & subt 3 in
num. Becomes 1 - 3/(x2+3):
Use formula ![]()
Some rational functions can be done by splitting the integrand into 2 fractions:
Ex: ![]()
![]()
CAS: Maple instructions for integration:
Sample: for ![]()
f:=sec(x)^3;
int(f,x); If, instead, you type Int(f,x), Maple will show you the unevaluated integral.
5.8 Evaluating Definite Integrals by Substitution
Either:
Method 1: Complete the indefinite integration, then apply the limits of integration
Ex: X20
Method 2: Write the limits of integration in terms of the new variable.
Ex: X22
Special cases:
Odd function with symmetric limits of integration: easy if you think about it.
Ex: ![]()
Some integrals correspond to geometric formulas:
let y = integrand, graph the function, view the geometric shape that is enclosed
Ex: int(sqrt(x2-4),x,0,2)
CAS: Maple instructions for integration:
Sample: for ![]()
f:=sec(x)^3;
int(f,x=0..Pi/3); If you type Int(f,x=0..Pi/3), Maple will show you the unevaluated integral.
7.1 Exponential and Logarithmic Functions
This lesson is essentially the precalculus approach to exponential and logarithmic functions.
Reminder: These are based on the
handling of rational exponents:
where b>0, b¹1
Exponential functions take form f(x) = bx, b>0 and b¹1: b is the "base"
Otherwise written as f(x) = exp(x)
Typical graphs, for bases that are greater than 1, or between 0 and 1.
Key points: (0,1), (1,base)
Log functions are defined as the inverse of the exponential functions:
So exp(log(x))=x and log(exp(x))=x for any base b, b>0, b¹1.
y = bx is equivalent to x = logb(y);
Typical graphs for base greater than 1, or between 0 and 1
Key points: (1,0), (base,1)
Theoretical problem: the precalc approach leaves these functions undefined for irrational numbers.
If true, the graphs should not be shown as continuous functions -- gaps for all irrational numbers
In fact, these are continuous, so we need to show how they apply to irrational exponents.
Solution: either
(1) Limit approach: sandwich a given irrational number between successively closer rationals
(2) [lesson 7.5] Totally different approach:
Define ln(x) as the integral of an appropriate continuous function [y=1/t, from 1 to x].
Natural exponential and natural log functions use the base e, where e is approximately 2.71828
So y = ex is equivalent to x = ln y of y>0 and x is any real number.
Definition of e =
(Compound interest
at 100% for 1 year compounded continuously)
See Thm 7.1.2: Exp and log functions are inverses
See Thm 7.1.3: Properties of logs: logb(ac) = logb(a) + logb(b) etc.
Proofs (exercise 46): let x = logb(a) and y = logb(c), solve for a and c, sub into left side
Tasks:
Expand in terms of sums, differences, and multiples of simpler logarithms
Ex: 
Rewrite an expression as a single logarithm [with a coefficient of 1]
Ex: ![]()
Change of base formula: ![]()
Ex: Use a calculator to find a decimal equivalent of log317
Know patterns of exp and log growth (and associated limits) from graphs of ex, e-x, ln x.
Applications
decibels, pH, radioactive decay, Richter scale, compound interest
Exercises are basically substitution tasks: see definitions before example 5 and in the exercise set.
7.2 Derivatives and Integrals Involving Log. Functions
Important derivatives (and corresponding integrals) from this lesson:
Dx ln(x) = 1 /x
Text shows proof for Dxlogbx = 1 / (x ln b)
[No need to learn formula for an alternate base b: use the change of base theorem]
Also, Dx ln(-x) = 1 / x (by chain rule), so Dx ln|x| = 1 / x
![]()
Ex: Find Dx ln(3x5)
1. Directly, using the formula above: Note how much cancels
2. By first applying properties of logs to expand the expression
Logarithmic Differentiation: Ex: like Example 6
Same principle as previous example would be nice, but it doesn't appear as a log expression. So
1. Let y = expressionb
2. Take natural log of both sides
3. Expand right side using properties of logs
4. Differentiate both sides implicitly: Note on left you get 1 / y * y'
5. Solve for y': You will end up multiplying the right side by the original expression.
7.3
Inverse
Functions
Definition of inverse functions
Informal: An inverse reverses the effect of the original function; Ex: f(x)=x3, g(x)=x1/3
Formal: f(g(x))=x for every x in the domain of g
g(f(x))=x for every x in the domain of f
Effects:
swap x and y values (in domain and range)
Domains and ranges are interchanged
reflect graph about diagonal line y=x
How to calculate inverses algebraically (simple cases).
Ex y = x/(x-2)
In tougher cases, we simply define the new function and invent a name for it (sin, arcsin)
Notation: Caution f-1¹ 1/f
When do inverse functions exist?
Inverse relations always exist; the question is, when is the inverse going to be a function?
Ex y=x^2 Inverse relation fails vertical line test.
Function must be one-to-one (abbrev. as 1-1) for the inverse to be a function
orig func never takes on the same value twice:
So orig func must pass horizontal line test
in symbols, f(x1) ¹ f(x2) whenever x1 ¹ x2
for proofs, If f(x1)=f(x2) then x1 = x2. ..[find f(x1) and f(x2), set equal, show x1=x2.]
Ex: Show f(x)=3x-2 is 1-1.
To show not 1-1: (1) conclusion doesn't follow, or (2) counterexample (y=x2 with y=4, x=±2)
Special case: functions that are strictly increasing or strictly decreasing
Domain must be a single interval
Technique to check 1-1 for some funcs: Is deriv always pos or always neg (and continuous)?
Ex: f(x) = x^3+x;
Caution: f(x) = tan(x) -- derivative sec2(x) is always pos., but domain consists of multiple intervals
Restricting domains
Algebra of inverses See def 2 pg 408: f-1(y)=x↔y=f(x) assuming domains and ranges are swapped
Note what happens to the function name f.
Ex: valid for y=sqrt(x) and y2=x if we restrict x to nonnegative reals.
Continuity: Either both f and g are continuous, or both discontinuous; obvious from graph.
Differentiability of inverse functions: See lab on derivative of inverse functions
Ex: find deriv of arctan
Ex: For f(x) =
x^3+x: find Dxf-1(30)
by inv func thm, without calculating
the inverse function.
x 0 1 2 3 4 f-1(x)
f(x) 0 2 10 30 68 x Dxf-1(30) = 1 / f'(f-1(30)) = 1/ f'(3) = 1/28
func inverse
Graphing functions and their inverses with graphing utilities
Technique: parametric mode
Derivatives and Integrals of Exp. Functions
Important derivatives (and corresponding integrals) from this lesson:
Dxex = ex Proved on lab for derivatives of inverse functions.
![]()
Ex: 
![]()
Derivatives and integrals involving the general exponential function bx:
Thm (not in book): bx = ex*ln b
Proof: let y = bx; take logs: ln y = x*ln b; take exp function of both sides.
So:
Note: multiply by ln b
Note: divide by ln b
Do not confuse xa (power rule) with ax (exponential function)
Differentiation of combinations (functions of x both in base and exponent): logarithmic differentiation.
Ex: Find derivative of (3x-1)cos(2x)
7.4
Graphs and Applications
Involving Logarithmic and Exponential Functions
Basic graphs of y=ex and y=ln(x)
Note from graphs:
Is domain or is range all positive? For ln(x), when are values (y) negative?
Increasing or decreasing functions?
Concave up or down?
Example: For f(x) = x2ln x, and for y=x2e-x, find: roots; y-intercept, limits at infinity, critical points (x,y),
maxima, minima, increasing & decreasing intervals, concavity & inflection points; sketch
Example: : For f(x) = e^(x^3-x) find: roots; y-intercept, limits at infinity, critical points (x,y),
maxima, minima, increasing & decreasing intervals, concavity & inflection points (get roots of 2nd derivative with calculator); sketch
Ex: Applications of integration
Logistics curves: See Ex 3 and X27: Form y = l / (1+Ae–kt)
You should be able to find:
Both horizontal asymptotes (as t approaches –∞ and ∞)
Initial value (let t = 0)
Rate of increase at any time t
Inflection point (Note: the y-coordinate is halfway between the asymptotes, i.e. at L/2, and the graph is symmetric about the inflection point)
Model: Rate of change of temperature is proportional to difference in temperature between object and its environment. Later, we’ll derive the equation.
This lesson starts with a temperature function, and we analyze it:
Find the rate of temperature change
Find average temperature over a period time
Find volume of a solid of rotation about y-axis, y = e^(-x^2), y=0, x=0, x=1, cylindrical method:
int(2*Pi*x*exp(-x^2),x=0..1) = Pi - Pi/e
7.5 L'Hôpital's Rule; Indeterminate Forms
Typical limit prob: find limit of f(x)/g(x) as x®a, where both f(a) and g(a) = 0
Techniques in 1st semester:
Factor and reduce:
Ex: (x-3) / (x2 - x - 6) as x®3 and as x®¥
Rationalize numerator or denominator:
Ex: [sqrt(2-t)-sqrt(2)] / t as t®0
Divide every term by 1/x^n
Ex: [5x3-2x+3] / [7-4x3] as x®∞ L'Hôpital requires repeated application
Interpret expression as a derivative and get the result by differentiation:
Ex: [esin x-1] / [x-p] as x®p This is deriv of esin x evaluated at x=p
Ex: [ln(1+x)] / x as x®0 This is deriv of ln(u) at 1, with "x" replacing "h"
Ex (similar): [ex - 1] / x as x®0 This is deriv of eu at 0, with "x" replacing "h"
Also see the example above for rationalizing the numerator.
Ex: (x+sin(3x)) / x as x®∞
Memorized forms
Ex: sin(x) / x as x®0
Ex: (1-cos(x)) / x as x®0
Add fractions, or split a single fraction into separate terms
Ex: 1/x – cos(x)/1 as x®0
Ex: (x+sin(3x)) / x as x®∞
Graphical approach (equivalently, table of values approach) to estimate the limit.
Caution: L'Hôpital's Rule is so powerful that students often forget all the other techniques.
But some problems can't be done with it (See #49-52, 58)
L'Hôpital's Rule: for 0/0 or ∞/∞, lim f(x)/g(x) = lim f '(x)/g'(x)
Other indeterminate forms:
¥/¥: handle like 0/0
¥*0: divide by the reciprocal of one of the factors to get 0/0 or ¥/¥
Ex: x2 csc(x2) as x®0
¥ - ¥: Combine in some way to get a single term (add frac; trig ident, etc.)
Ex: 1 / ln(x) - 1/ (x-1) as x®1 After adding, requires repeated applic.
Indeterminate powers: Use logarithmic technique
00
Why is the form 00 indeterminate? x0 ®1 as x®0+, but 0x ® 0 as x®0+ Inconsistent.
Ex: x sin x as x®0
¥0
Ex: x1/x as x®¥
1¥
Ex: Calculate e as either [1+1/t]t as t®¥ or [1+t]1/t as t®0
Doesn't work if form doesn't hold
Repeated applications are often necessary
7.6 Log Functions from the Integral Point of View
Definition: ln
x =
, x>0
Problem with integrating 1/x by power rule: division by zero.
But the area under the curve
exists -- so the integral
exists (for any
positive number b).
The definition interprets the area as a function of the upper limit of integration x.
Apply First Fund Thm of Calc, Part 2 (pg 367) to get
Dx(ln x) = 1/x;
Dx(ln u)=1/u Dxu
Properties of logs follow from the derivative, along with
ln 1 = 0 (because the upper and lower limits of integration are the same)
ln e =1 (we take this as the definition of the number e)
For properties, ln x and ln ax have the same deriv, so they differ by const;
ln ax = ln x + C; let x=1 to get C = ln a
We define the inverse of ln(x) to be exp(x):
Motivation: log func is 1-1, has inverse, so: Def of exp(x) = y Û ln y = x [is equiv to, or "iff"]
Canx eqns are exp(ln x) = x and …
To show exp(x)=ex, simplify ln(ex) = x ln e = x; take exp( ) of both sides.
Rewrite canx eqns in terms of ex instead of exp(x)
Familiar laws of exponents: Prove e^(x+y) = e^x e^y by taking logs of both sides.
Deriv of e^x is e^x. Prove by inverse func procedure
Proof that this e is the same as the precalculus e:
Find derivative of ln(1+x) when x=0 by the formula (result is 1)
and
also by the definition of the derivative: ![]()
Equate the results, and
solve for
= e
Problem 6 deals with the midpoint rule. Click here for an illustration of how to do this with EXCEL.
Alternate procedure for initial value problems:
If dy/dx = f(x) and y(x0) = y0, then sol. is y(x) = y0 + int(f(x), x0..x)
I prefer that students continue to do indef integration and then sub the initial condition to find C -- this procedure has a wider application
Derivatives of integrals with functions as limits of integration. How to use the chain rule?
7.7 Derivatives and Integrals Involving Inverse Trig Functions
Definitions of the six inverse trig functions:
They are the inverses of the basic trig functions
Since the basic trig functions are periodic, we have to restrict their domains to get 1-1 functions
Where is the basic func 1-1? That's where the inverse is defined
sin & tan: -p/2 to p/2
cos & cot: 0 to p
sec & csc: include the 1st quadrant &
for the sec, quadrant 2 (like cos)
problem: diff and int formulas for the secant include abs value.
(to see why, do the inverse function procedure for y=sec-1x with y in Q2)
for the csc, quadrant 4 (like sin)
(cotangent and cosecant are not addressed n the text, because they are not needed)
Some texts picks 3rd quadrant for secant to avoid an abs value in the diff and int formulas)
--
Differentiation formulas for arcsin, arctan, arcsec [the cofunctions are the negatives of these]
Know integration formulas in terms of u2 and a2, resulting in:
,
(both with coeff of 1/a); and
(no coeff of 1/a).
See integration formulas 68, 77, 87 in BOB (back of book)
Sample interals: algebraic: int(t^2/sqrt(4-t^6),t) u=t^3, a=2
algebraic: int((x+9)/(x^2+9),x) What approach? Split in two;
In first, u-sub with u=x^2+9; in second, arctan with u=x, a=3
exponential: int (exp(2x)/sqrt(3-exp(4x)),x)
u=exp(2x), a=sqrt(3); result is arcsec
trig: int ( sin(x)/(2+cos2x),x)
u=cos(x), a=sqrt(2)
For X88 draw and label triangles for α = arctan(x) and β = arctan(y); solve for α and β; use trig identity for tan (α+β); take arctan of both sides
7.8. Hyperbolic Functions and Hanging Cables
Defs of sinh and cosh in terms of exp(x).
Be able to write recognize expr for sinh(ln x) etc
Defs of others in terms of sinh and cosh
Know graphs of sinh, cosh; see tanh: Refer to graphsof exp(x) and exp(-x)
Note odd/even behavior is same as for trig.
Note hanging chain
Basic identites: Know cosh2x - sinh2x = 1 Be able to prove.
Why hyperbolic? point on curve of unit hyperbola is sinh, cosh because cosh2x - sinh2x = 1