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Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

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Presentation on theme: "Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)"— Presentation transcript:

1 Derivatives Part A

2 Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.) The derivative of a constant multiplied by a variable is the constant. 3.) The derivative of a variable to a power is found by multiplying the coefficient by the power and then subtracting 1 from the power.

3 Review of Basic Rules f(x)=sin(X) f`(x)=cos(X) f(x)=g(x) ± h(x)f`(x)= g`(x) ± h`(x)   f(x)=cos(X)  f`(x)= - sin(X) 4.) The derivative of the trig function Sine is Cosine. 5.) The derivative of the trig function Cosine is Sine. 6.) If you have terms that are added or subtracted you may find the derivative of each term separately and the combine the derivative.

4 The Product Rule : When you have expressions multiplied, you assign the temporary variable “u” to one expression and the temporary variable “v” to the other. Then you find the derivative of “U” and call it “du”. Find the derivative of “V” and call it “dv”. Insert the values into the product rule and evaluate it.

5 Product Rule Review: f(x) = 3x 2 ( x 3 + 2 ) f `(x) = uv` + vu` f `(x) = ( )( ) + ( )( ) u v v` = 3x 2 u` = 6x 3x 2 x 3 + 2 9x 4 + 6x 4 + 12x 15x 4 + 12x 3x 2 6x

6 The Quotient Rule : When you have expressions divided, you assign the temporary variable “u” to the numerator’s expression and the temporary variable “v” to the denominator’s expression. Then you find the derivative of “U” and call it “du”. Find the derivative of “V” and call it “dv”. Insert the values into the quotient rule and evaluate it.

7 Quotient Rule: f(x) = x 2 + 3 x + 5 f`(x) = vu`- uv` v2v2 ( )( ) – ( )( ) ( ) 2 2x v` =1 u` = f `(x) = 2x 2 + 10x x 2 +10x + 25 x 2 + 10x - 3 u v x 2 + 3 2x 1 x + 5 x2x2 3 - - x 2 +10x + 25 = – f `(x) = x + 5 - -

8 Notation: f(x)  This notation tells the user that the answer depends on how the variable x changes. We say that the answer is a “Function” of x. This notation indicates we are getting the 1 st derivative of y based on x. Means: Find the 1 st derivative of y based on the variable u Means: Find the 1 st derivative of u based on the variable x

9 The Chain Rule

10 The Chain Rule states that if you have a composite function, a function within a function, you can break down the function to simplify it. Many times we can rewrite a complicated function as a composite, which will allow us to use our basic derivative rules to find the derivative.

11 Examples of breaking down a composite function.

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14 Steps for using the CHAIN RULE

15 Examples of using the CHAIN RULE

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