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Derivative Rules
Calculus - Derivatives
More Calculus Lessons
Many applications of calculus require us to deduce facts about a function f from the information concerning its derivatives. Since f ’(x) represents the slope of the curve y = f(x) at the point (x, f(x)), it tells us the direction in which the curve proceeds at each point.
Example:
Find where the function f(x) = x^{3} – 12x + 1 is increasing and where it is decreasing.
Solution:
Step 1: Find the derivative of f
f ’(x) = 3x^{2} – 12 = 3(x^{2} – 4) = 3(x – 2)(x + 2)
Step 2: Set f ’(x) = 0 to get the critical numbers.
f ’(x) = 3(x – 2)(x + 2) = 0
x = 2, –2
Step 3: Set up intervals whose endpoints are the critical numbers and determine the sign of f ‘(x) for each of the intervals. Use the increasing/decreasing test to determine whether f(x) is increasing or decreasing for each interval.
Interval |
x – 2 |
x + 2 |
f ’(x) |
f |
x < – 2 |
– |
– |
+ |
Increasing on |
–2 < x < 2 |
– |
+ |
– |
Decreasing on (–2, 2) |
x > 2 |
+ |
+ |
+ |
Increasing on |
Suppose that c is a critical number of a continuous function f.
Example:
Find the local maximum and minimum values of the function f(x) = x^{4} – 2x^{2} + 3
Solution:
Step 1: Find the derivative of f
f ’(x) = 4x^{3} – 4x = 4x(x^{2} – 1) = 4x(x – 1)(x + 1)
Step 2: Set f’(x) = 0 to get the critical numbers
f ’(x) = 4x(x – 1)(x + 1) = 0
x = –1, 0, 1
Step 3: Set up intervals whose endpoints are the critical numbers and determine the sign of f’(x) for each of the intervals.
Interval |
4x |
x –1 |
x + 1 |
f ’(x) |
x < –1 |
– |
– |
– |
– |
–1 < x < 0 |
– |
– |
+ |
+ |
0 < x < 1 |
+ |
– |
+ |
– |
x > 1 |
+ |
+ |
+ |
+ |
Step 4: Use the first derivative test to find the local maximum and minimum values.
f ’(x) goes from negative to positive at x = –1, the First Derivative Test tells us that there is a local minimum at x = –1.
f (–1) = 2 is the local minimum value.
f ’(x) goes from positive to negative at x = 0, the First Derivative Test tells us that there is a local maximum at x = 0.
f (0) = 3 is the local maximum value.
f ’(x) goes from negative to positive at x = 1, the First Derivative Test tells us that there is a local minimum at x = 1.
f (1) = 2 is the local minimum value.
We can also use the Second Derivative Test to determine maximum or minimum values.
The Second Derivative Test
Suppose f ’’ is continuous near c,
Example:
Use the Second Derivative Test to find the local maximum and minimum values of the function f(x) = x^{4} – 2x^{2} + 3
Solution:
Step 1: Find the derivative of f
f ’(x) = 4x^{3} – 4x = 4x(x^{2} –1) = 4x(x – 1)(x + 1)
Step 2: Set f’(x) = 0 to get the critical numbers
f ’(x) = 4x(x – 1)(x + 1) = 0
x = –1, 0, 1
Step 3: Find the second derivative
f ’’(x) = 12x^{2} – 4
Step 4: Evaluate f ‘’at the critical numbers
f ’’(–1) = 8 > 0, so f (–1) = 2 is the local minimum value.
f ’’(0) = – 4 < 0, so f (0) = 2 is the local maximum value.
f ’’(1) = 8 > 0, so f (1) = 2 is the local minimum value.
Relative Maxima and Minima (Local Maxima and Minima)
Finding relative maxima and minima of a function can be done by looking at a graph of the function. A relative maximum is a point that is higher than the points directly beside it on both sides, and a relative minimum is a point that is lower than the points directly beside it on both sides. Relative maxima and minima are important points in curve sketching, and they can be found by either the first or the second derivative test.
The first derivative test is a way to find if a critical point of a continuous function is a relative minimum or maximum. Simply, if the first derivative is negative to the left of the critical point, and positive to the right of it, it is a relative minimum. If the first derivative test finds the first derivative is positive to the left of the critical point, and negative to the right of it, the critical point is a relative maximum.
Increasing and Decreasing Functions
Determine the intervals for which a function is increasing and/or decreasing by using the first derivative.
The First Derivative Test to find the Relative Extrema.
Increasing/Decreasing and Local Max and Mins using First Derivative Test.
Inflection points are points on the graph where the concavity changes. A positive second derivative means a function is concave up, and a negative second derivative means the function is concave down. These inflection points are places where the second derivative is zero, and the function changes from concave up to concave down or vice versa.
Determine the concavity of a function on an interval.
This video provides an example of how to determine the intervals for which a function is concave up and concave down as well as how to determine points of inflection.
Concavity, Inflection Points and Second Derivatives
Examples of using the second derivative to determine where a function is concave up or concave down.
The second derivative test is useful when trying to find a relative maximum or minimum if a function has a first derivative that is zero at a certain point. Since the first derivative test fails at this point, the point is an inflection point. The second derivative test relies on the sign of the second derivative at that point. If it is positive, the point is a relative minimum, and if it is negative, the point is a relative maximum.
Use the second derivative test to determine the relative extrema.
This video provides an example of how to use the second derivative test to determine relative extrema of a function.
Finding Local Maximums/Minimums - Second Derivative Test.
This video shows how to use the second derivative test to find local maximums and local minimums.
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