Calculating the Derivative

NOTATIONS FOR THE DERIVATIVE The derivative of y = f ( x ) may be written in any of the following ways You need to become very familiar with the Rules for Derivatives Summary on page 722. You must understand the meaning and use of each notation used on that page and be able to apply it correctly to each term or function for which you are finding the derivative. CONSTANT RULE If f (x) = k, where k is any real number, then f ' ( x ) = 0. (The derivative of a constant is 0.) Think about this from a logical standpoint. The first derivative yields the rate of change. If k is a constant, how much does it change over any period of time? It does not change because if it did, it would not be a constant. So logically the derivative of a constant will always be zero. We will mainly use the first two notations.

POWER RULE If f ( x ) = x n for any real number n, then f ' ( x ) = n x n – 1. (The derivative of f (x) = x n is found by multiplying the exponent n and decreasing the exponent by 1.) Example: f ( x ) = x 8 f ' ( x ) = 8 x 8 – 1 f ' ( x ) = 8 x 7 You will need to be able to work with exponents of different types such as exponents that are fractions both positive and negative, improper fractions, and negative integers. It is necessary that students be familiar with rules learned in algebra and to be able to use these rules even in reverse mode. In algebra it is taught that negative exponents are not desirable. One of the rules learned is that. However there are times when negative exponents are more preferable. An example would be to find the first derivative of. As of now, we have no rule for finding the derivative with x in the denominator. But if we rewrite the problem as f ( x ) = x – 4, the power rule can be used: f ' ( x ) = – 4 x – 4 – 1 = – 4 x – 5.

CONSTANT TIMES A FUNCTION Let k be a real number, then D x [ k f ( x ) ] = k f ' ( x ). (The derivative of a constant times a function is the constant times the derivative of the function.) SUM OR DIFFERENCE RULE If f ( x ) = u ( x )  v ( x ).and if u ' ( x ) and v ' ( x ) exist, then f ' ( x ) = u ' ( x )  v ' ( x ) (The derivative of a sum or difference of functions is the sum or difference of the derivative.) This rule is what allows us to find the derivative of a function that has more than one term. Example: f ( x ) = 5 x 7 f ' ( x ) = 5 ( 7 ) x 6 f ' ( x ) = 35 x 6

PRODUCT RULE If f ( x ) = u ( x )  v ( x ) and if u ' ( x ) and v ' ( x ) exist, then f ' ( x ) = u ( x )  v ' ( x ) + v ( x )  u ' ( x ). (The derivative of a product of two functions is the first function times the derivative of he second, plus the second function times the derivative of the first.) Example: Find the derivative of y = ( 7x – 4 )( 3x + 5 ) Solution: Let u ( x ) = ( 7x 7x – 4 ) and v ( x ) = ( 3x 3x + 5 ), then u ' ( x ) = 7 and v ' ( x ) = 3 f ' ( x ) = 3( 7x 7x – 4 ) + 7( 3x 3x + 5 ) f ' ( x ) = 21 x – x + 35 f ' ( x ) = 42 x + 23 Note: I do not want you to multiply the ( 7x – 4 )( 3x + 5 ) and then find the derivative. Even though this would be correct, I want you to demonstrate that you understand the product rule. Substituting this into the power rule and using the commutative rule that states ab = ba we obtain the following:

QUOTIENT RULE If f ( x ) = u ( x ) / v ( x ), if u ' ( x ) and v ' ( x ) exist, and v ' ( x ) ≠ 0, then (The derivative of a quotient is the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator.) Example: Find the derivative of Solution: Let u ( x ) = ( 4x 4x + 7 ) and v ( x ) = ( 5x 5x – 2 ), then u ' ( x ) = 4 and v ' ( x ) = 5 Substituting these into the quotient rule we obtain the following:

CHAIN RULE If y is a function of u, say y = f ( u ), and if u is a function of x, say u = g ( x ), then y = f ( u ) = f [ g ( x ) ] and, One way to remember the chain rule is to pretend that dy/du and du/dx are fractions, with du “canceling out.” Find dy/dx if y = ( 5 x 2  7 x ) 4 Solution: Let y = u 4, and u = 5 x 2  7 x. Then = 4 u 3  (10 x  7) and = 4 ( 5 x 2  7x 7x ) 3 (10 x  7)

= 4 ( 10 x  7 ) ( 5 x 2  7 x ) 3 I prefer the order Some people prefer using CHAIN RULE (ALTRNATIVE FORM) If y = f [ g ( x ) ], then (To find the derivative of f [ g ( x ) ], find the derivative of f ( x ), replace each x with g ( x ), and then multiply the result by the derivative of g ( x ).) = f ' [ g ( x ) ]·g ' ( x ). Find dy/dx if y = 7( 3 x x – 8 ) 4 Solution: If the problem was f ( x ) = 7 x 4, the derivative would be f ' ( x ) = 7( 4 ) x 3 = 28 x 3.3. If the problem was g ( x ) = 3 x x – 8, the derivative would be g ' ( x ) = 6 x + 5.

So if we put these ideas together we can find the derivative. f ' ( x ) = 7( 4 )( 6 x + 5 ) ( 3 x x – 8 ) 3 f ' ( x ) = 28( 6 x + 5 ) ( 3 x x – 8 ) 3 Generic steps that I use: 1. Multiply the coefficient by the exponent 2. Leave a space 3. Copy parenthesis as is 4. Reduce the exponent by one 5. In the space, write the derivative of the parenthesis See page 698 example 5. I prefer to write g ' ( x ) in front rather than at the back of the answer.

Find the derivative of y = 3 ( 7 x 2 – 3 x + 4 ) 4 ( 2 x x – 8 ) 3. Using a combination of the product rule and the alternative form of the chain rule we can find the derivative. If the exponents 4 and 3 did not exist, this problem would be the product rule. Because they are in the problem, we will also use the chain rule. Solution: Let u ( x ) = ( 7 x 2 – 3 x + 4 ) 4 and v ( x ) = ( 2 x x – 8 ) 3, then u ' ( x ) = 4 ( 14 x – 3 )( 7 x 2 – 3 x + 4 ) 3 and v ' ( x ) =3( 4 x + 7 )( 2 x x – 8 ) 2 Substituting into the product rule would give y ' = 3( 4 )( 14 x – 3 )( 7 x 2 – 3 x + 4 ) 3 ( 2 x x – 8 ) 3 + 3( 3 )( 4 x x 2 – 3 x + 4 ) 4 ( 2 x x – 8 ) 2 I keep u ( x ) and v ( x ) in the same order as they appear in the problem. It is easier to keep track of what is happening. The derivative has been found but if you want full credit for the problem you must do the algebra required to get the answer in final form.

Next factor the greatest common factor (GCF). y ' = 3 ( 7 x 2 – 3 x + 4 ) 3 ( 2 x x – 8 ) 2 [ 4 ( 14 x – 3 )( 2 x x – 8 ) + 3 ( 4 x x 2 – 3 x + 4 )] Multiply the factors and add like terms as seen on the next slide. This is scratch work that I want to see on a test question but not in the problem itself. This is why I want you to use a blue book for tests. Use the left side of the booklet to show scratch work. Scratch work is a necessary part of working a problem but it is not part of the solution. y ' = 3 ( 7 x 2 – 3 x + 4 ) 3 ( 2 x x – 8 ) 2 ( 196 x x 2 – 547 x ) Answer:

Step 1. 4 ( 14 x – 3 )( 2 x x – 8 ) ( 56 x – 12 )( 2 x x – 8 ) 56 x ( 2 x x – 8 ) – 12 ( 2 x x – 8 ) 112 x x 2 – 448 x – 24 x 2 – 84 x x x 2 – 532 x + 96 Step x x 2 – 532 x x x 2 – 15 x x x 2 – 547 x Step 2. 3 ( 4 x + 7 )( 7 x 2 – 3 x + 4 ) ( 12 x + 21 )( 7 x 2 – 3 x + 4 ) 12 x ( 7 x 2 – 3 x + 4 ) + 21 ( 7 x 2 – 3 x + 4 ) 84 x 3 – 36 x x x 2 – 63 x + 84 x x 2 – 15 x + 84