Calculus I
Prelims
ME11S3 | Joshua Miguel Riveral | SEM 1 2024
● Now that we can plug anything into
functions, we can plug one function
Functions
in as the input of another function.
● A function is a way of matching up ● This is called composition. The
one set of numbers with another. composition of function f with
The first set of numbers is called function g is written f o g. This
the domain. means to plug g into f like this:
● For each of the numbers in the ● (f o g)(x) = f(g(x))
domain, the function assigns
exactly one number from the other
Domain
set, the range
● For example, the domain of the ● When an expression is used to
function could be the set of describe a function f(x), it is
numbers {1, 4, 9, 25, 100}, and the convenient to think of the domain
range could be {1, 2, 3, 5, 10}. as the set of all numbers that can
Suppose the function takes 1 to 1, be substituted into the expression
4 to 2, 9 to 3, 25 to 5, and 100 to and get a meaningful output.
10. This could be illustrated by the ● This set is called the domain. The
following: range of the function is the set of
all possible numbers produced by
evaluating f at the numbers in its
domain.
Practice Problem
What is the domain of f(x) = 3/x-2
● We must never let the
Practice Problem
denominator be zero, so x cannot
Find the value of g(3) if g(x) = x2+2 be 2. Therefore, the domain of this
function consists of all real
numbers except 2.
● The prohibition against even roots
(like square roots) of negative
numbers is less severe. An even
root of a negative number is an
imaginary number. Useful
mathematics can be done with
imaginary numbers.
Practice Problem ● However, for the sake of simplicity,
we will avoid them.
Find the value of h(4) if h(t) = t3-2t2+5
Practice Problem
What is the domain of g(x) =
Composition of functions
1
Calculus I
Prelims
ME11S3 | Joshua Miguel Riveral | SEM 1 2024
Limits and Continuity
● The notion of a limit is the single
most important underlying
concept upon which calculus is
built.
● Limits can be illustrated using
graphs and tables of values.
How to write in interval notation?
● We can’t talk about f(x) at x = 2
because of the unshaded circle on
its graph. But, we can talk about
what happens close to 2. The
values of the function at x-values
close to 2 are listed in the table.
Personal Note
A parenthesis is used when we intend to
NOT include a point, whereas a square
bracket is used when we intend TO
include a point.
Exponents
● Exponents frequently arise in
calculations throughout calculus.
If a is a positive real number and
n is a positive integer (that is, n =
1, 2, 3, ...), then an means Limits and Continuity
“multiply the base a by itself n ● The mathematical shorthand for
times.”
this is = 5, which is read
● Symbolically, an=a x a x a…a as “the limit as x approaches 2 of
● The exponential function f(x) = ex f(x) is 5.”
and the natural logarithm g(x) = ● The utility of limits lies in the fact
ln(x) “undo” each other when that f need not be defined at the
composed. value a in order to have a limit as x
approaches a.
2
Calculus I
Prelims
ME11S3 | Joshua Miguel Riveral | SEM 1 2024
● We can also approach points from divided, or composed, it works also
either the left or from the right. for every function considered.
● The little minus in lim means that Example
we approach x = 1 using numbers
less than (to the left) of x = 1. As
we approach x = 1 from the
left-hand side, we slide up the
graph through y-values that
approach 4. Similarly, the plus in
lim means “approach from the
right.”
Practice Problem
● Here we see two functions
approaching 2. It suggests what
the graph is doing on either side of
2. The limits technically do not
exist or they are discontinuous
since they are infinite.
Evaluating Limits algebraically
● It is not necessary to have the
graph of a function to evaluate its
limits. For instance, if f is
Practice Problem
continuous at a, then its limit as x
approaches a is simply f(a).
● Technically, this works only with
functions that are polynomials like
4x5+10x3-7, roots like x1/2, rational
functions like 3x-5/2x3+x2+1,
trigonometric functions like sin(x),
and transcendental functions like
ln(x), and ex.
● Because this works for any
combination of these functions
added, subtracted, multiplied,
3
Calculus I
Prelims
ME11S3 | Joshua Miguel Riveral | SEM 1 2024
Practice Problem
Practice Problem
Practice Problem
4
Calculus I
Midterms
ME11S3 | Joshua Miguel Riveral | SEM 1 2024
Differentiation by definition (Four-step
rule) Rules of differentiation
● Let f(x) be the given function. 1. Constant rule - The derivative
The derivative of f(x) is given by of the constant is always equal
to zero.
2. Power rule - Let f(x)=x^n where n
is any natural number be f’(x) =
nx^n-1
● The derivative of the function f(x) 3. Constant multiple rule - For any
can be solved using the Four-step function f and any constant c, d/dx
rule. [cf(x)] = c d/dx [f(x)].
● 1st step: Determine f(x+Δx) by 4. Addition and subtraction rule -
changing all x by x + Δx on the f(x) = g(x) ± h(x) —--> f’(x) = g’(x) ±
given function. h’(x)
● 2nd step: Subtract f(x) from 5. Product rule - f(x)=g(x)×h(x) →
f(x+Δx). f’(x) = udv + vdu where u=g(x)
● 3rd step: Divide the expression and v=h(x).
both sides by Δx. 6. Quotient rule - f(x)=g(x) / h(x) →
● 4th step: Apply lim Δx→ 0 on the f’(x) = vdu - udv / v² where u=g(x)
third step. and v=h(x) and v≠0.
Derivative notations Chain rule
● Lagrange’s notation: f’ ● If the function g is differentiable
● Leibniz's notation: ((d/dx)f(x)) at x and the function f is
dy/dx differentiable at g(x) , then the
composite function.
Practice Problem ● f o g is differentiable at x, and (f o
g)′ (x) = f′(g(x)) ∙ g′(x). Or simply
Find all f’ of f(x) = 2x f’(g(x)) = f′(g(x)) ∙ g′(x), is the
derivative of the outer most
function f times the derivative of
the inner most function (g(x)).
● In Leibniz notation, if y is a
differentiable function of u, and u is
differentiable function of x, then y
is a differentiable function of x and,
(dy/dx) = (dy/du) ∙ (du/dx).
Practice Problem
f(x) = (3x+1)²
1
Calculus I
Midterms
ME11S3 | Joshua Miguel Riveral | SEM 1 2024
Practice Problem
Differentiation of Exponential and
f(x) = √13x²-5x+8
Logarithmic Functions
● This module focuses on finding
derivatives of functions involving
exponents and logarithms.
● Formulas:
Practice Problem
f(x) = ³√(4x+1/x^5)
● Chain Rule Application: When
differentiating composite functions
with exponential or logarithmic
components, the chain rule is
essential.
Practice Problem
y=3e^5x
Personal Note
If the differentiation gets very tedious,
don't overcomplicate things. Let the rules
of differentiation do the work one-by-one.
Another worth noting is that the solutions
I've got is very detailed so if you got the
correct answer anyway then don't check
the solution anymore.
2
Calculus I
Midterms
ME11S3 | Joshua Miguel Riveral | SEM 1 2024
Implicit differentiation
Practice Problem
y= 2a^5x ● It would be quite difficult to
re-arrange this so y was given
explicitly as a function of x
● We could, perhaps, given values of
x use the expression to work out the
values of y and thereby draw a
graph. In general, even if this is
possible, it would be difficult.
● A function given in this way is said to
be implicit. In these modules, we will
study on how to differentiate
functions given in this form.
Practice Problem
x3+y3=4
Practice Problem
y=e9x-6+a9x-6
Practice Problem
y= x2y3+x3y2
3
Calculus I
Midterms
ME11S3 | Joshua Miguel Riveral | SEM 1 2024
Practice Problem
(x-y)2=x+y-1
Hyperbolic functions
Inverse Hyperbolic functions
Trigonometric functions
Inverse trigonometric functions
Practice Problem
y = cos(x2-3x+2)
4
Calculus I
Midterms
ME11S3 | Joshua Miguel Riveral | SEM 1 2024
Practice Problem
y = cot(x+y)
Practice Problem
y = arccsc(3x+2)
Practice Problem
y = 4arccos x - 10arctan x
5
Calculus I
Midterms
ME11S3 | Joshua Miguel Riveral | SEM 1 2024
Practice Problem
y = (3x-2) cosh(3x-2)
Practice Problem
y = csch(5x)
Practice Problem
y = arccoth(8x-4)
6
Calculus I
Midterms
ME11S3 | Joshua Miguel Riveral | SEM 1 2024
Personal Note
While tons of memorization is needed. We
must look for patterns to recognize and
familiarize to memorize less formulas. An
example of this are formulas in hyperbolic
functions that are the same for
trigonometric functions except only sech,
csch, and tanh are negatives and the rest
are positives.
Parametric Differentiation
● Let f and g be two real–valued
functions of the real variable of t .
Then for every number t in the
domain common to f and g ,
the curve C is set of all points
(x, y) such that x = f(t) and y =
g(t) are called parametric
Practice Problem equations of the curve C , and t
is called a parameter .
y = (9x-3) arccosh(9x-3)
● Thus, if dy/dt = 0 and dx/dt ≠ 0 at
some point on the curve C, there is
a horizontal tangent at that point.
● If dy/dt = 0 and dx/dt = 0 , further
investigation is needed to
determine the inclination of the
tangent line, if it exists.
Practice Problem
Find y’ and y”
x = 3t, y = 2t2
Practice Problem
7
Calculus I
Midterms
ME11S3 | Joshua Miguel Riveral | SEM 1 2024
2
y = x3-2x+1
x = 1 – t , y = 1 + t.
Practice Problem
x = t2et, y = t ln(t).
Practice Problem
y=sin(x) at x = pi / 4
Slope as the Derivative of a Function
● Suppose the function f is
continuous at x1 . The tangent line
to the graph of f at the point P(x0,
f(x0)).
● The equation of a tangent line with
slope m and passing through (a, b)
is y = m(x – a) + b .
● The normal line to graph at a given
point is the line perpendicular to
the tangent line at that point.
● The normal line has a slope m =
–1 / f′(x0).
Practice Problem
8
Calculus I
Midterms
ME11S3 | Joshua Miguel Riveral | SEM 1 2024
Practice Problem
Find the x-coordinate where the tangent line
to y = ln(x) is horizontal
Practice Problem
y = x2 parallel to the line y = 4x+3
9
Calculus I
Finals
ME11S3 | Joshua Miguel Riveral | SEM 1 2024
Find all critical points of f (x) = sin x + cos x
Critical Points on point [0,2π].
● The domain of f (x) is restricted to
● Points on the graph of a function the closed interval [0,2π].
where the derivative is zero or the
derivative does not exist are
important to consider in many
application problems of the
derivative.
● The point (x, f (x)) is called a
critical point of f (x) if x is in the
domain of the function and either
f'(x) = 0 or f'(x) does not exist.
● If (x, f(x)) is a critical point of f(x), ●
then x is called a critical number of
f(x). The geometric interpretation of FDT & SDT
what is taking place at a critical
point is that the tangent line is ● Derivative tests are used to find
either horizontal, vertical, or does whether or not a critical point has a
not exist at that point on the curve. maxima, minima, or even an
inflection. There are 2 ways to find
it out.
Practice Problem ● The first derivative test (FDT) is a
Find all critical points of f (x) = x4 – 8x2 method for determining whether a
● Because f (x) is a polynomial function has a local maximum or
function, its domain is all real minimum at a critical point. It
numbers. examines a function's monotonic
properties, or whether it's
increasing or decreasing, at a
specific point in its domain.
○ If f’(x) changes from
positive to negative, it has a
maximum
○ If f’(x) changes from
negative to positive, it has a
minimum
○ If f’(x) does not change, it
has neither a maximum nor
a minimum
● The second derivative test (SDT) is
much easier. Simply taking the
● second derivative of a function.
● Hence, the critical points of f (x) ○ If f”(x)<0, it has a maximum
are (–2,–16), (0,0), and (2,–16), ○ If f”(x)>0, it has a minimum
and the critical numbers are x = ○ If f”(x)=0, go back to FDT
–2, 0, and 2.
Extreme Value Theorem
● An important application of critical
points is in determining possible
maximum and minimum values of
a function on certain intervals.
● The Extreme Value Theorem
guarantees both a maximum and
1
Calculus I
Finals
ME11S3 | Joshua Miguel Riveral | SEM 1 2024
minimum value for a function under Since f′′(x)>0, there is a minimum
certain conditions. It states the at x=2.
following: If a function f (x) is
continuous on a closed interval ● Step 4: Find the value of the
[a,b], then f (x) has both a function at x=2.
maximum and minimum value on
[a,b]. f(2)=22−4(2)+3
● The procedure for applying the f(2)=4−8+3
Extreme Value Theorem is to first f(2)=−1
establish that the function is
continuous on the closed interval. So, the minimum value is −1 at
● The next step is to determine all x=2.
critical points in the given interval
and evaluate the function at these
critical points and at the endpoints
of the interval. The largest function
value from the previous step is the
maximum value, and the smallest
function value is the minimum
value of the function on the given
interval.
●
Practice Problem
Find the maximum and minimum values of Personal Note
f (x) = x4 – 3x3 – 1 on [–2,2].
First derivative is where you get the critical
point/s. Set f’(x)=0 to solve. The second
derivative is used to find the minimum or
maximum of a function. The value you get
doesn't matter as long as you find if it has
a maximum (-) or a minimum (+). If you
get 0 value. Solve it using the FDT.
●
Point of inflection (POI)
Practice Problem ● An inflection point of a function
𝑓(𝑥) is where the curve changes
Find the extreme values of f(x)=x2−4x+3 concavity—that is, it shifts from
being concave up (shaped like a
● Step 1: Find the first derivative. cup: ∪) to concave down
f′(x)=2x−4 (shaped like a cap: ∩) or vice
versa. At an inflection point, the
● Step 2: Set the derivative equal to second derivative 𝑓'′(𝑥) typically
zero.
changes sign.
2x−4=0
2x=4
● x=2 this is a critical point. Practice Problem
Find the inflection points of f(x)=x3
● Step 3: Use the second derivative
test to determine if it's a maximum
or minimum.
f′′(x)=2
2
Calculus I
Finals
ME11S3 | Joshua Miguel Riveral | SEM 1 2024
examples where extreme values
are applicable:
Practice Problem
A rectangular garden is to be constructed
against the side of a garage. The gardener
has 100ft of fencing and will construct a
3-sided fence; the side of the garage will
form the fourth side. What dimension will
give the garden of greatest area?
Practice Problem
Determine the concavity of f (x) = sin x +
cos x on [0,2π] to identify any points of
inflection of f (x).
Practice Problem
Personal Note
A rectangular plot of land is to be fenced in
Second derivative is where you get the using two kinds of fencing. Two opposite
critical point/s. Set f”(x)=0 to solve. If there sides will use heavy-duty fencing selling for
are 2 or more critical points. Use FDT to $3 a foot, while the remaining two sides will
find which critical point/s to use. use standard fencing selling for $2 a foot.
What are the dimensions of the rectangular
plot of greatest area can be fenced in at a
cost of $6,000?
Applications of the Extreme Values
(Optimization)
● Extreme values (maximum and ● Let x and y be the dimensions
minimum) are important concepts and A the area.
not just in mathematics, but also in
a variety of real-life applications. In
many practical situations, we are
concerned with optimizing a
particular outcome—whether that
means maximizing efficiency,
minimizing costs, or finding the
best possible solutions to a
problem. Here are some few The cost condition is
3
Calculus I
Finals
ME11S3 | Joshua Miguel Riveral | SEM 1 2024
Applications of the Extreme Values
3(2x) + 2(2y) = 6000 (Related Rates)
6x+ 4y = 6000
[3x + 2y = 3000] (1/2) ● Extreme values (maximum and
minimum) are important concepts
(3/2)x + y = 1500
not just in mathematics, but also in
y = 1500 – (3/2)x a variety of real-life applications. In
many practical situations, we are
The function is A = xy concerned with optimizing a
particular outcome—whether that
Substituting the value of y to the means maximizing efficiency,
given function A . minimizing costs, or finding the
best possible solutions to a
A = x[1500 – (3/2)x] problem. Here are some few
A = 1500x – (3/2)x2 examples where extreme values
are applicable:
Differentiate.
A’ = 1500 – 3x Practice Problem
0 = 1500-3x A 13-foot ladder leans against a wall. The
x = 500 foot of the ladder begins to slide away from
Using SDT the wall at the rate of 1 foot per minute.
A” = -3 , f”(x)<0 When the foot is 5 feet from the wall, at
what rate is the top of the ladder falling?
x = 500, and has a relative
maximum value.
To solve for y , substitute x =
500.
y = 1500 – (3/2)(500)
y = 1500 – (1500/2)
y = 750
For the Area:
A = (500)(750)
A = 375,000 ft2
Therefore, the area is greatest
when x = 500 ft and y = 750
ft .
Practice Problem
Personal Note A light is hung 15 feet above a straight
horizontal path. If a man 6 feet tall is
Clearly understand the problem first and walking away from the light at the rate of 5
illustrate it. Find out what formula to use. ft/sec, how fast is his shadow lengthening?
Once that’s over, find the extreme values
then substitute it in the formula.
4
Calculus I
Finals
ME11S3 | Joshua Miguel Riveral | SEM 1 2024
Exercise
Two cars, one going due east at the rate of
90 km/hr and the other going due south at
the rate of 60 km/hr, are traveling toward
the intersection of two roads. At what rate
are the cars approaching each other at the
instant when the first car is 0.2 km and the
second car is 0.15 km from the
intersection?
Practice Problem Partial Differentiation
Water is flowing at the rate of 2 cubic ● Partial differentiation is the process
meters per minute into a tank in the form of of finding the derivative of a
an inverted cone having an altitude of 16 function with respect to one
meters and a radius of 4 meters. How fast is variable while keeping all other
the water level rising when the water is 5 variables constant. It is commonly
meters deep? used in functions of two or more
variables, such as 𝑓(𝑥,𝑦) or 𝑔(𝑥,𝑦,𝑧).
Practice Problem
Find Zx and Zy of the function f(x,y) = x2+y2
Practice Problem
Find Zx and Zy of the function f(x,y) =
x2+sin(xy)
5
Calculus I
Finals
ME11S3 | Joshua Miguel Riveral | SEM 1 2024
Practice Problem
Find Zx, Zy, and Zz of the function f(x,y) =
x2z+exysin(z)
