1 2 3 4 5 6 7 8 9 10

0:0010 lessons across the whole video55:57

You will be able to

Evaluate limits by factoring, canceling, and direct substitution
Differentiate polynomial, rational, exponential, logarithmic, and composite functions using the power, product, and chain rules
Find antiderivatives with the power rule and u-substitution
Find the slope and equation of a tangent line using implicit differentiation
Solve related rates problems for a changing cylinder volume
Locate intervals of increase, decrease, concavity, and extrema using derivative sign analysis

Good to know first

Factoring quadratics and difference of squares
Basic exponent rules, including negative and fractional exponents
Understanding of functions, graphs, and slope
Familiarity with the basic definitions of derivatives and integrals

Syllabus

10 lessons

1
Limits by Factoring and Substitution, then the Power Rule Free
0:00–5:005 min of video
You will learn to evaluate a limit that gives 0 divided by 0 by factoring and cancelling, then finishing with direct substitution. You will also learn the power rule and use it to differentiate a sum of powers of x, including negative and fractional exponents.
2 formulas2 exercises3 questions
2
Continuity of Piecewise Functions, and Differentiating e^u, ln u, and Products Beyond Pass
5:00–10:275 min of video
You will learn how to find the constant that makes a piecewise function continuous by setting its two pieces equal at the switching point. You will also learn the derivative rules for e^u, ln u, and the product of two functions, and use them together.
3 exercises4 questions
3
Find antiderivatives with the power rule Beyond Pass
10:27–14:094 min of video
You will learn the power rule for integration: add one to the exponent, then divide by the new exponent. You will use it after simplifying a fraction so that every term is a simple power of x.
3 exercises4 questions
4
Find a tangent line with implicit differentiation Beyond Pass
14:09–21:488 min of video
You learn how to find the slope of a curve at a point when y is not written as a function of x, using implicit differentiation. Then you turn that slope and the point into the equation of the tangent line, in point-slope and standard form.
3 exercises4 questions
5
Limit-Definition Derivatives and Integrals by u-Substitution Beyond Pass
21:48–25:424 min of video
You will learn to recognize a limit that is secretly the definition of a derivative, and to evaluate integrals by u-substitution, replacing an inner function with u and its differential with du. Both skills rely on spotting the right pattern before calculating.
3 exercises3 questions
6
Related Rates: The Filling Cylinder Beyond Pass
25:42–32:096 min of video
You will learn how to solve a related rates problem for a cylinder: write the volume formula, decide which quantities are constants and which are variables, differentiate with respect to time, and substitute the given values. The chapter ends with the rate of change of the volume, 27π cubic feet per minute.
2 formulas2 exercises3 questions
7
Critical Points, Increasing Intervals, and Maxima Beyond Pass
32:09–40:278 min of video
You learn to find critical points by setting the first derivative equal to zero, then use a sign chart to decide where a function is increasing or decreasing. You also use the sign change at a critical point to find a local maximum and its value.
2 exercises4 questions
8
Average value of a function and the chain rule Beyond Pass
40:27–47:317 min of video
You will learn how to find the average value of a function on an interval using a definite integral, and how to differentiate a function inside a power with the chain rule. Both ideas are worked through with the video's own examples.
2 formulas3 exercises3 questions
9
Evaluating Limits Numerically and by Simplifying Complex Fractions Beyond Pass
47:31–52:085 min of video
You learn two ways to find a limit: plug in numbers closer and closer to the target to see where the values settle, and clear the small fractions inside a complex fraction so you can substitute directly. Both methods are checked on the same problem, the limit as x approaches 4 of (1/x − 1/4)/(x − 4).
1 formula3 exercises3 questions
10
Intervals of Concavity and Inflection Points Beyond Pass
52:08–55:574 min of video
You learn how the sign of the second derivative tells you whether a graph is concave up or concave down, and how to find inflection points where the concavity changes. You then apply this to find the intervals where f(x) = x^3 - 6x^2 + 5x + 1 is concave downward.
3 exercises3 questions

Free preview · Lesson 1 in full

Limits by Factoring and Substitution, then the Power Rule

You will learn to evaluate a limit that gives 0 divided by 0 by factoring and cancelling, then finishing with direct substitution. You will also learn the power rule and use it to differentiate a sum of powers of x, including negative and fractional exponents.

0:00 to 5:00 in the video

Scratchpad

Draw and write next to the video.

The idea

Start every limit by trying direct substitution: plug the number in and see what you get. In problem 1, plugging x = 3 into the bottom gives 3 squared minus 9, which is 0. A fraction with 0 on the bottom cannot be evaluated this way, so you need another method.

The fix is to factor. The bottom, x squared minus 9, is a difference of perfect squares, so it factors as (x + 3)(x - 3). The top, x squared + 2x - 15, is a trinomial: you need two numbers that multiply to -15 and add to 2. Those are 5 and -3, so the top is (x + 5)(x - 3).

Now the factor (x - 3) appears on both the top and the bottom. Cancelling it is allowed because we only care about x near 3, not at 3 itself. What is left is the limit as x approaches 3 of (x + 5)/(x + 3). This new expression has no 0 on the bottom, so direct substitution works: (3 + 5)/(3 + 3) = 8/6 = 4/3. The answer is 4/3.

The power rule handles derivatives of powers of x. If you have x to the power n, its derivative is n times x to the power n minus 1. For x to the 4th, that is 4x cubed. The rule works for any number n, including negative and fractional ones.

To use the power rule on a messy expression, first rewrite every term as x raised to a power. 3/x becomes 3x to the power -1, and the square root of x becomes x to the power one half. Then differentiate each term separately and add the results, using the sum rule.

Problem 2 gives the derivative of x^6 + 3/x - sqrt(x). The x^6 term becomes 6x^5. The 3x^(-1) term becomes 3(-1)x^(-2) = -3x^(-2). The x^(1/2) term becomes (1/2)x^(-1/2). Then move the negative exponents back to the bottom of fractions: -3/x^2 and 1/(2 sqrt(x)). The final answer is 6x^5 - 3/x^2 + 1/(2 sqrt(x)).

Key points

Always try direct substitution first; only factor when you get 0 on the bottom. 0:29
Factor the top and bottom, then cancel the common factor, which is allowed because x only approaches 3. 0:43
After cancelling, substitute the number into the simpler expression to get the limit. 1:54
Power rule: the derivative of x^n is n x^(n-1), and it works for negative and fractional n. 2:21
Rewrite roots and fractions as powers of x before differentiating, then move negative exponents back to the bottom. 3:22
Differentiate each term of a sum separately. 2:54

Formulas

\frac{d}{d x} (x^{n}) = n x^{n - 1}

The derivative of x to the power n is n times x to the power n minus one.

$x$: the variable
$n$: a constant exponent (any real number)
$n - 1$: the new exponent, one less than the old one

\lim_{x \to 3} \frac{x^{2} + 2 x - 15}{x^{2} - 9} = \lim_{x \to 3} \frac{x + 5}{x + 3} = \frac{4}{3}

The limit as x approaches 3 of the original fraction equals the limit of the simplified fraction, which equals 4 over 3.

$\lim_{x \to 3}$: the value the expression approaches as x gets close to 3
$x^{2} + 2 x - 15$: the numerator, which factors as (x+5)(x-3)
$x^{2} - 9$: the denominator, which factors as (x+3)(x-3)

Worked example

Evaluate the limit as x approaches 3 of (x^2 + 2x - 15)/(x^2 - 9).

Direct substitution gives 0 on the bottom (3^2 - 9 = 0), so factor instead.
Bottom: x^2 - 9 = (x + 3)(x - 3).
Top: x^2 + 2x - 15 = (x + 5)(x - 3), since 5 and -3 multiply to -15 and add to 2.
Cancel the common factor (x - 3), leaving the limit of (x + 5)/(x + 3).
Substitute x = 3: (3 + 5)/(3 + 3) = 8/6 = 4/3.

4/3

Differentiate x^6 + 3/x - sqrt(x).

Rewrite: x^6 + 3x^(-1) - x^(1/2).
Differentiate x^6: 6x^5.
Differentiate 3x^(-1): 3(-1)x^(-2) = -3x^(-2).
Differentiate -x^(1/2): -(1/2)x^(-1/2). (The minus sign from the original expression is kept.)
Move negative exponents to the bottom: -3/x^2 and -1/(2 sqrt(x)) with the sign handled as shown.

6x^5 - 3/x^2 + 1/(2 sqrt(x))

Watch out

Getting 0 on the bottom does not mean the limit is 0 or undefined; it means you must factor and simplify first. 0:29
When differentiating 3/x, do not write the derivative of 3 as 0 and forget the x; rewrite it as 3x^(-1) first. 1:51
Subtract 1 from the exponent, not divide: the power of x changes from n to n minus 1. 2:21

Practice

Calculate

Evaluate the limit as x approaches 2 of (x^2 - 4)/(x - 2).

Hint

Factor x^2 - 4 as a difference of squares, cancel, then substitute.

Solution

Direct substitution gives 0/0.
Factor: x^2 - 4 = (x + 2)(x - 2).
Cancel (x - 2): the limit becomes the limit of (x + 2).
Substitute x = 2: 2 + 2 = 4.

Calculate

Find the derivative of x^3 + 2/x evaluated at x = 1.

Hint

Rewrite 2/x as 2x^(-1), differentiate each term, then substitute x = 1.

Solution

Rewrite: x^3 + 2x^(-1).
Differentiate: 3x^2 + 2(-1)x^(-2) = 3x^2 - 2x^(-2).
Substitute x = 1: 3(1) - 2(1) = 1.

On your scratchpad

Draw a fraction (x^2 - 9)/(x^2 + 2x - 3) style example on your scratchpad: write the top and bottom, factor each one, circle the matching factor, and cross it out.

Quiz

What is the first step in evaluating the limit of (x^2 - 9)/(x - 3) as x approaches 3?
1. Differentiate the top and bottom
2. Try direct substitution, which gives 0/0, so factor
3. Answer 0 immediately
4. Multiply by the conjugate
Show the answer
B. Try direct substitution, which gives 0/0, so factor · Direct substitution gives 0/0 here, so the video factors the top and bottom to simplify.
What is the derivative of x^4 using the power rule?
1. 4x^3
2. x^3
3. 4x^4
4. 4x^5
Show the answer
A. 4x^3 · The power rule gives n x^(n-1) = 4x^(4-1) = 4x^3.
What is the derivative of 3/x?
1. -3/x^2
2. 3/x^2
3. -3x^2
4. 3x^(-2)/x
Show the answer
A. -3/x^2 · Rewrite 3/x as 3x^(-1). The derivative is 3(-1)x^(-2) = -3x^(-2) = -3/x^2.

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Lessons 2–10

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Made by Beyond from the video “Calculus 1 Final Exam Review” by The Organic Chemistry Tutor. The video plays in YouTube's own player. Watch it on YouTube · All lessons

10 lessons

Limits by Factoring and Substitution, then the Power Rule Free

Continuity of Piecewise Functions, and Differentiating e^u, ln u, and Products Beyond Pass

Find antiderivatives with the power rule Beyond Pass

Find a tangent line with implicit differentiation Beyond Pass

Limit-Definition Derivatives and Integrals by u-Substitution Beyond Pass

Related Rates: The Filling Cylinder Beyond Pass

Critical Points, Increasing Intervals, and Maxima Beyond Pass

Average value of a function and the chain rule Beyond Pass

Evaluating Limits Numerically and by Simplifying Complex Fractions Beyond Pass

Intervals of Concavity and Inflection Points Beyond Pass