CHAPTER 2 LESSON 3
For both the product rule and the quotient there is a recognizable format that will allow you tho know which rule you will need to use and for each rule there is a formula to memorize and put into effect to find the derivative.
The product rule:
The product rule is recognized as F(x)G(x)
The formula for solving with the product rule is
D/Dx [F(x)G(x)] = F(x) Gprime(x) + G(x) Fprime(x)
Example: (3x-2x^2) (5+4x)
First I take the derivative e of each so that when I plug in I already know what the derivatives are to be plugged in
D/Dx [3x-2x^2]= 3-4x
Therefore F prime= 3-4x
D/Dx [5+4x]= 4
Therefore G prime= 4
Now you plug into the formula
Therefore
3x-2x^2(4)+ 5+4x (3-4x)
Distribute
12x^2x-8x^2+15-20x+12x-16
Simplify
-24x^2 + 4x +15
Because this equation cannot be simplified any further
Dx= -24x^2+4x+15
The quotient rule:
The quotient rule is recognized as F(x)/G(x)
The formula for solving with the quotient rule is
D/Dx [F(x)/G(x)] = G(x) Fprime(x)- F(x) Gprime(x)/ [G(x)]^2
Example: 5x-2/x^2-1
First I take the derivative of each so that when I plug in I already know what the derivatives are to be plugged in
D/dx [ 5x-2] = 5
Therefore F prime= 5
D/dx [x^2 – 1] = 2x
Therefore G prime = 2x
Now you plug into the formula
Therefore
(x^2+1)(5)-[(5x-2)(2x)]/(x^2+1)^2
Distribute
5x^2+5-[10x^2-4x]/(x^2+1)^2
Distribute the negative
5x^2+5-10x^2+4x/(x^2+1)^2
Simplify
-5x^2 + 4x+5/ (x^2+1)^2
Because this equation cannot be simplified any further
DX= -5x^2 + 4x+5/ (x^2+1)^2
Some things to remember:
**don’t forget that when solving with the quotient rule the negative must be distributed to the entire product of f(x)G prime(X)
** don’t forget which order F(X)G prime(X) and G(X)F prime(X) go in for each equation because it DOES matter.
CHAPTER 2 LESSON 4
The chain rule could loosely be defined as an order of operations that is used when solving composite functions.
A very important key concept to remember when using the chain rule is to work from the outside to the inside.
The “Formula” for the chain rule is
d/dx [fg(x)] = f ’(g(x)) (g’(x))
The most common procedure to using the chain rule is
• First, take the derivative of the outside
• Second, recopy just the inside
• Finally, multiply by the derivative of the inside
Example:
Square root of 3x^2-X+1
= (3x^2-X+1)^1/2
First, take the derivative of the outside:
½(________)^-1/2
Second, recopy just the inside
½(3x^2-X+1)^-1/2
Finally, multiply by the derivative of the inside
½(3x^2-X+1)^-1/2 (6x-1)
Simplify
6x-1/2(3x^2-x+1)^1/2
CHAPTER 2 LESSON 5
The key thing to remember when taking implicit derivatives is that the steps are the same as taking a regular derivative.
“with respect to” is a phrase you will see often when dealing with implicit derivatives.
The phrase “with respect to” gives you a clue as to what you will be solving for
When you see “d/dx” you also must be able to recognize that this is a symbol for “with respect to x”
When you take a derivative of anything besides x you will write it in the asterisk spot of the following example: D*/Dx
In other words, the top letter next to the d is the variable you are taking the derivative of
For example:
With respect to y: dy/dx
With respect to r: dr/dx
With respect to s: ds/dx
To solve a problem you must follow three general steps:
First take the derivative
Second gather all d*/dx
Third solve for d*/dx
Example:::
Sqrtxy= x^2y+1
(xy)1/2=x^2y+1
X^ 1/2 y^ ½ = x^2 y +1
First take the derivative
X^1/2 (1/2y^-1/2 dy/dx) + y^1/2 (1/2x^-1/2)= x^2 (1dy/dx+y(2x))
1/2x^1/2y^-1/2 dy/dx + 1/2x ^-1/2 y^1/2 = x^2 dy/dx + 2xy
Second gather all d*/dx
1/2x^1/2y^-1/2 dy/dx- X^2 dy/dx= 2xy - 1/2x^-1/2y^1y^1/2
Third solve for d*/dx
Dy/dx = 2xy -1/2x^-1/2y^1/2 / ½ x^1/2 y^-1/2 – x^2
Example with trig:
4cosx siny=1
4[cosx(cosy dy/dx)+ siny (-sinx)] = 0
4cosxx cosy dy/dx – 4 siny sin x =0
4 cosx cosy dy/dx= 4siny sinx
Dy/dx= 4 siny sinx / 4 cosx cos y
Dy/dx= tan y tan x
We also learned how to take the second derivative when dealing with implicit derivatives
The steps to solving the second derivative of an implicit derivative are:
Take the first derivative
Solve for dy/dx
Take the derivative of dy/dx (this will be d^2y/dx^2)
Plug in the result of when you solved for dy/dx into the d^2y/dx^2 equation for dy/dx
Solve again
**** sometimes when you solve the numerator will turn out to be your original equation, if so, plug in what the original equation equaled and simplify.
If not then d^2y/dx^2 will be what you found when you “solved again”
Example:
x^2 + y^2 = 25
Take the first derivative
2x + 2y dy/dx = 0
Solve for dy/dx
Dy/dx= -2x/2y therefore -x/y
Take the derivative of dy/dx
- Y(1)- [ x(1dy/dx)]/y^2
- Y-x dy/dx / y^2
Plug in the result of when you solved for dy/dx into the d^2y/dx^2 equation for dy/dx
- Y-x ( -x/y)/ y^2
Solve again
- Y^2 + x^2/y/y^2/1
-y^2 +x^2/ y^3
**** remember that sometimes when you solve the numerator will turn out to be your original equation, if so, plug in what the original equation equaled and simplify.
Therefore:
-25/y^3
CHAPTER 2 LESSON 6
There are a set of steps to follow when solving a related rate problem but first you need to know how to recognize a related rate problem.
A related rate problem is one that involves something per something.
I.E. Gallons per minute
Cubic feet per second
Your key word is PER
Keep in mind that although there are set steps for working related rate problems but each problem is unique in its own method of solution.
To further explain the steps are guidelines but the specific application is dependent upon each problem.
The steps to solving these problems are as follows.
First, identify all given quantities and quantities to be determined. Also make a sketch and be sure to label the quantities.
Second, you must write an equation involving the variables whose rates of change either are given or are to be determined
Third, you must use the chain rule implicitly differentiate both sides of the equation with respect to time (T)
Finally, you must substitute all known variables into the resulting equation and solve for the desired rate of change
Be sure to print and study your Geometric formulas
Example:
A Pebble is dropped into a calm pond causing ripples in the form of concentric circles. The radius of the outer circle is increasing at a rate of 1 foot PER second when the radius is 4 ft at what rate is the total area (A) of the disturbed water changing?
#1 dr/dt= 1ft/sec
R=4 ft
dA/dt= ?
#2 A= pi r ^2
#3 dA/dt= pi{2r dr/dt}
#4 dA/dt= 2pi (4)(1)
Therefore 8 pi ft/sec
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