{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecture12

# lecture12 - Lecture 12 1 Lecture 12(Sec 3.1 3.2 Basic Rules...

This preview shows pages 1–9. Sign up to view the full content.

Lecture 12 1 Lecture 12: (Sec. 3.1, 3.2) Basic Rules of Differentiation; Product and Quotient Rules Recall: if y = f ( x ) is differentiable, then f ( x ) = d dx [ f ( x )] = Some Basic Rules for Finding Derivatives 1) Derivative of a Constant : If f ( x ) = c for any constant c , then f ( x ) = ex. Find d dx ( 2)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Lecture 12 2 NOTE: If f ( x ) = x , then f ( x ) = parenleftbigg d dx ( x ) = parenrightbigg a54 a45 a63 a27 Proof: 2) The (Simple) Power Rule : If f ( x ) = x n , then f ( x ) = where n is any real number.
Lecture 12 3 Find: ex. d dx ( x 165 ) ex. d dx ( 1 x 5 ) ex. d dx ( 5 x )

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Lecture 12 4 For the following rules, we assume that f and g are differentiable functions of x : 3) The Derivative of a Constant Multiple of a function : If g ( x ) = cf ( x ), then g ( x ) = where c is any constant. We can also say: d dx [ cf ( x )] = Proof :
Lecture 12 5 ex. Find f ( x ) for each function: 1) f ( x ) = x 3 2) f ( x ) = 2 5 x 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Lecture 12 6 4) The Sum Rule d dx [ f ( x ) ± g ( x )] = Proof, p. 164 ex. Find f ( x ) if f ( x ) = 3 x 6 + 2 5 x 2 2 x 3 2
Lecture 12 7 ex. 1) Find d dx ( x + 3) 2) Find d dx (3 x ) ex. Find f (0) if f ( x ) = x 2 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Lecture 12 8 The Product Rule ex. Differentiate: h ( x ) = ( x 2 + 2)(3 x 1) NOTE: d dx [ f ( x ) g ( x )] negationslash =
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern