Chapter 5 section one-
covers taking the derivative of a natural log.
For completing this task there is a simple formula to follow.
this formula is
D/dx ln(u)= 1/u X (du/dx)
This formula is read as the derivative of natural log u is found by putting one over what is in parenthesis multiplied by the derivative of what had been in parenthesis.
Therefore you would plug into the formula
Simplify
and
solve to get your result.
DO NOT FORGET THAT YOUR DERIVATIVE RULES SUCH AS THE CHAIN RULE THE PRODUCT RULE AND THE QUOTIENT RULE STILL APPLY.
Some examples of chapter five section one include:
'd/dx ln(2x) = 1/2x (2)
d/dx ln(2x)= 2/2x
Therefore ln(2x)= 1/x
d/dc ln (x^2+1)= 1/x^2+1 (2x)
Therefore ln(x^2+1)= 2x/x^2+1
d/dx x ln (x) = x(1/x (1)) + lnx (1)
d/dx x ln (x)= x(1/x)+ln x
d/dx x ln (x)= x/x+lnx
Therefore d/dx x ln (x)= 1 + lnx
d/dx (ln(x))^3= 3(ln x)^2 (1/x) (1)
d/dx (ln(x))^3= 3(ln x)^2 (1/x)
Therefore d/dx (ln(x))^3= 3(ln x)^2/ x
Chapter five section two
covers taking the derivative of an ecponential function.
For completing this task there is also a simple formula to follow.
the formula for solving these problems is
D/dx e^u = e^u X (du/dx)
This formula is read as the derivative of the exponential function e is found by recopying what the problem is asking to take the derivative of and multiplying it by the derivative of the exponent.
Therefore when trying to solve these problem too you would plug into the formula
Simplify
and
solve to get your result.
AGAIN!
DO NOT FORGET THAT YOUR DERIVATIVE RULES SUCH AS THE CHAIN RULE THE PRODUCT RULE AND THE QUOTIENT RULE STILL APPLY.
Some examples of these type problems include
d/dx e^2x-1 = e^2x-1 (2)
Therefore d/dx e^2x-1 = 2e^2x-1
d/dx e^-3/x= e^-3/x (-3x^-1)
d/dx e^-3/x= e^-3/x (3x^-2)
d/dx e^-3/x= e^-3/x (3/x^2)
Therefore d/dx e^-3/x= 3e^-3/x / x^2
Chapter five section four
covered taking the derivative for exponents other than e
For completing this task there are a few simple formulas to follow.
one of the formula for solving these problems is
D/dx A^u = A^u(ln (a))(du/dx)
This formula is read as the derivative of a number raised to a variable as a power is found by multiplying what the problem is asking to take the derivative of by ln of the number being raised to the variable and then multiplying all of that by the derivative of the exponent.
Therefore when trying to solve these problem too you would plug into the formula
Simplify
and
solve to get your result.
Some examples of these type problems include
d/dx 2^x= 2^xln2 (1)
Therefore the derivative of 2^x is 2^x ln 2
d/dx 2^3x= 2^3x ln 2 (3)
Threfore the derivative of d/dx 2^3x is 3 (2^3xln2)
d/dx 7 ^x^2= 7^x^2ln7 (2x)
Therefore the derivative of d/dx 7 ^x^2 is 2x(7^x^2ln7)
The other formula for solving these problems is
d/dx logaU= 1/U(ln(a)) X (du/dx)
This formula is read as the derivative of a log is found by placing what the problem is asking for in asking for the derivative of the log under one then multiplying it by the derivative of what the problem is asking for in asking for the derivative of the log.
Therefore when trying to solve these problem too you would plug into the formula
Simplify
and
solve to get your result.
Some examples of these type problems include
d/dx log cosx = 1/cosx(ln10) (-sin x)
d/dx log cosx = -sinx/ cosx(ln10)
Therefore the derivative of log cos x= -tanx/ ln10
d/dx Log5 x^3+4x= 1/x^3+4x(ln5) (3x^2+4)
Therefore d/dx Log5 x^3+4x= 3x^2+4/ (x^3+4x)(ln5)
STEPS TO USING THE SECOND DERIVATIVE TEST:
Step #1:
take the first derivative set equal to zero and solve. ((keep in mind that this step will not always be necessary but can always be performed anyway. This step is not necessary when one looking for concavity or points of inflection.))
Step #2:
take the second derivative set = 0 and solve
Step #3:
Check differentiability
Step #4:
set up intervals
Step #5:
pick numbers on the interval and plug them into the SECOND derivative
Step #6:
if POSITIVE: the function is concave up on that interval
if NEGATIVE: the function is concave down on that interval
Step #7:
If there is a change it is a point of inflection ((This step only needs to be done when the problem is asking for the points of inflection specifically))
Example #1:
Determine the open intervals on which the graph is concave up or concave down:
F(x)= 6/ x^2+3
Step #1:
take the first derivative set equal to zero and solve. (( But this step is not necessary because the problem is looking for concavity.))
Step #2:
take the second derivative set = 0 and solve
(x^2+3)(0)-[(6)(2x)]/ (x^2+3)^2
F'(x)= -12x/ (x^2+3)^2
(x^2+3)^2(-12)-[(-12x)(2(x^2+3)(2x))]/ (x^2+3)^2
(x^2+3)(-12)+48x^2
-12x^2-36+48x^2
F"(x)= -36x^2-36=0
-36x^2=36
x^2= 1
x= +/- 1
Step #3:
Check differentiability
DIFFERENTIABLE?
yes.
Step #4:
set up intervals
(-infinity,-1)U(-1,1)U(1, infinity)
Step #5:
pick numbers on the interval and plug them into the SECOND derivative
F"(-2)= 36(-2)^2-36/ (-2^2+3)^2 = POSITIVE
F"(0)= 36(0)^2-36/ (0^2+3)^2 = NEGATIVE
F"(2)= 36(2)^2-36/ (2^2+3)^2= POSITIVE
Step #6:
if POSITIVE: the function is concave up on that interval
Therefore the function is concave up on the intervals (-infinity, -1)U(1,Infinity)
if NEGATIVE: the function is concave down on that interval
Therefore the function is concave down on the interval (-1,1)
Step #7:
If there is a change it is a point of inflection.
This step only needs to be done when the problem is asking for the points of inflection specifically therefore this step is not necessary for this problem because it is not asking for the points of inflection.
THE SHORTCUT TO THE FIRST DERIVATIVE TEST
Steps to the shortcut for the first derivative test:
Step #1:
take the derivative, set equal to zero, and solve
Step #2:
Check to see if differentiable
Step #3:
Plug in critical points into the second derivative
Step #4
If POSITIVE there is a min
If NEGATIVE there is a max
If 0 the test is false
(((NOTICE THAT FOR THE SHORTCUT WHETHER THE PLUG IN IS POSITIVE OR NEGATIVE HAS A REVERSED MEANING)))
Example:
Step #1:
take the derivative, set equal to zero, and solve
-15x^4+15x^2=0
15x^2(-x^2+1)
x= 0, 1,-1
Step #2:
Check to see if differentiable
differentiable?
yes.
Step #3:
Plug in critical points into the second derivative
F"(X)=60x^3 + 30x
-60(0)^3+30(0)=0
-60(-1)^3+30(-1)=POSITIVE
-60(1)^3+30(1)=NEGATIVE
Step #4
If POSITIVE there is a min
Therefore, at x= -1 there is a min
If NEGATIVE there is a max
Therefore, at x= 1 there is a max
If 0 the test is false
Therefore, at x=0 the test fails
Horizontal asymptotes and curve sketching
To find a horizontal asymptote there are three rules to follow
these rules apply to the degree of the leading coefficients of the polynomials in both the numerator and denominator
These rules are:
* If the top degree of the leading coefficient is equal to the bottom degree of the leading coefficient then you take the coefficient of the highest degree in the numerator and put it over the coefficient of the highest degree in the denominator
Example:
What is the horizontal asymptote of
2x^2+5 / 3x^2+1
because the degrees of the leading coefficients are equivalent there is a horizontal asymptote at 2/3
*If the top degree of the leading coefficient is greater than bottom degree of the leading coefficient then there is a horizontal asymptote at - infinity , infinity. In other words there is no horizontal asymptote.
Example:
What is the horizontal asymptote of
2x^2+5 / 3x+1
because the degree of the leading coefficient in the top is greater than the leading coefficient in the bottom the result is
-infinity, infinity and therefore there are no horizontal asymptotes.
*If the top degree of the leading coefficient is less than bottom degree of the leading coefficient then there is a horizontal asymptote at 0.
Example:
What is the horizontal asymptote of
2x+5 / 3x^2+1
because the degree of the leading coefficient in the top is less than the leading coefficient in the bottom the horizontal asymptote is 0.
For curve sketching problems there are five steps to follow
these steps are:
#1
find the domain and find the range
The domain is found by setting the bottom equal to zero, solving for x, and setting up intervals
The range is found by finding the horizontal asymptotes and setting up intervals
#2
Find the x intercepts, y intercepts, vertical asymptotes, and horizontal asymptotes.
X intercepts are found by setting the original equal to zero and solving for x to get the point (result,0)
Y intercepts are found by plugging in 0 for x and solving to get the point (0,result)
Vertical asymptotes and horizontal asymptotes have already been found at this point
your vertical asymptote is the x= from when you found the domain before you set up your intervals
your horizontal asymptote is the y= from when you found the range before you set up your intervals
#3
Do the first derivative test by
Taking the derivative and setting it equal to zero
Solve for x
check differentiability
Set up intervals (-infinity, pt)U(pt, infinity)
Plug in value on the interval into the derivative and solve
((if you get a positive number the function is increasing, If you get a negative number the function is decreasing))
(( A max is increasing then decreasing, A min is decreasing then increasing))
#4
Do second derivative test
take the derivative of the first derivative
set it equal to zero
solve for x
check differentiability
set up intervals (-infinity, pt)U(pt,infinity)
plug into the second derivative and solve
((if you get a positive number the function is concave up, If you get a negative number the function is concave down))
((any point on the interval where a number switches from concave up to concave down or vise versa there is a point of inflection))
#5
Plot all important information on a sketched graph.
IMPORTANT INFORMATION INCLUDES:
domain intervals
range intervals
x intercepts
y intercepts
vertical asymptotes
horizontal asymptotes
whether the intervals of the first derivative test are increasing or decreasing
where there is a max or a min
whether the intervals of the second derivative test are concave up or concave down
Example:
y= 2(x^2-9)/x^2-4
#1
find the domain and find the range
The domain:
x^2-4=0
x^2=4
x = +/- 2
(-infinity,-2)(-2,2)(2,infinity)
The range is found by finding the horizontal asymptotes and setting up intervals
y= 2
(-infinity,2)(2,infinity)
#2
X intercepts are found by setting the original equal to zero and solving for x to get the point (result,0)
2(x^2-9)=0 therefore: x=+/-3
Y intercepts are found by plugging in 0 for x and solving to get the point (0,result)
2(0^2-9)/0^2-4
2(-9)/-4
-18/-4
9/2
(0,9/2)
your vertical asymptote is the x= from when you found the domain before you set up your intervals
your horizontal asymptote is the y= from when you found the range before you set up your intervals
#3
Do the first derivative test by
Taking the derivative and setting it equal to zero
x^2-4(4x)-[2x^2-18(2x)]/(2^2-4)^2
20x/(x^2-4)^2
Solve for x
20x=0
x=0
check differentiability
vertical asymptotes at +/- 2
Set up intervals (-infinity, pt)U(pt, infinity)
(-infinity,-2)(-2,0)(0,2)(2,infinity)
Plug in value on the interval into the derivative and solve
f(-3) negative therefore decreasing
f(-1) negative therefore decreasing
f(1) positive therefore increasing
f(3) positive therefore increasing
MIN @ x=0
((if you get a positive number the function is increasing, If you get a negative number the function is decreasing))
(( A max is increasing then decreasing, A min is decreasing then increasing))
#4
Do second derivative test
take the derivative of the first derivative
-20(3x^2+4)/(x^2-4)^3
set it equal to zero
-20(3x^2+14)=0
solve for x
x=sqrt -4/3 therefore does not apply
check differentiability
vertical asymptotes at +/-2
set up intervals (-infinity, pt)U(pt,infinity)
(-infinity, -2)(-2,2)(2, infinity)
plug into the second derivative and solve
f(-3) negative concave down
f(0) positive concave up
f(3) negative concave down
THE STEPS TO OPTIMIZATION: (are also listed on page 219 of our text book)
#1
list what is given to you in the problem
#2
list the primary equation
#3
List the secondary equation
#4
Solve for y in the secondary equation
#5
plug y= into the primary equation for y
#6
set = 0 and solve
#7
perform the first derivative test
meaning:
take the derivative
set = 0
solve for x
set up intervals
plug in for F(x) between intervals into the derivative
determine if result is positive or negative and thereby increasing or decreasing respectivley on the intervals
therefore enabling you to determine whther x= is a max or a min
EXAMPLE:
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. what dimensions will produce a maximum volume?
#1
given: SA=108in^2 SA= x^2+4xy
Primary Equation: V= x^2y
Secondary equation: x^2+4xy=108
4xy=108-x^2
y= 108-x^2/ 4x
Plug in: V= X^2(108-x^2/4x)
V= 108x-x^3/4
27x-1/4x^3
first derivative test:
27-3/4x^2=0
-3/4x^2= -27
x^2=36
x= + - 6
Set up intervals
(0,6) U (6,sqrt108)
F(1) is positive and thereby increasing
F(7) is negative and thereby decreasing
Therefore X=6 is a maximum
Therefore V= LXWXH
V= 6X6XH
6^2 x y = 108
y= 108/ 36
y=3
V= 6X6X3
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