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**Unformatted text preview: **MATH 8, SECTION 1, WEEK 5 - RECITATION NOTES TA: PADRAIC BARTLETT Abstract. These are the notes from Monday, Oct. 25ths lecture, where we begin our discussion of the derivative. 1. Random Question Question 1.1. Can you find a function f : [0 , 1] [0 , 1] such that f (0) = 0 ,f (1) = 1 , f is continuous, and everywhere f has a derivative, its derivative is 0? 2. Differentiation: Definitions Definition 2.1. For a function f defined on some neighborhood ( a- ,a + ), we say that f is differentiable at a iff the limit lim h f ( a + h )- f ( a ) ( a + h )- a exists. If it does, denote this limit as f ( a ); we will often call this value the deriv- ative of f at a . The derivative has a number of interpretations as physical phenomena, which we will discuss in class on Wednesday; first, however, we will simply calculate a few derivatives to show how to attack these kinds of problems. 3. Differentiation: Examples Lemma 3.1. The derivative of f ( x ) = 1 /x at any point a 6 = 0 is- 1 /a 2 . Proof. Pick any point a 6 = 0 in R : then, a direct calculation of the limit tells us that lim h f ( a + h )- f ( a ) ( a + h )- a = lim h 1 a + h- 1 a h = lim h a ( a ) ( a + h )- a + h ( a ) ( a + h ) h = lim h a- a- h ( a ) ( a + h ) h = lim h - h ( h ) ( a ) ( a + h ) = lim h - 1 ( a ) ( a + h ) 1 2 TA: PADRAIC BARTLETT But this is just the quotient of a pair of polynomials, the denominator of which is nonzero as h 0. Because polynomials are continuous, we then know that we can pass the limit through the quotient operation, and just evaluate the numerator and denominators limits separately. Consequently, we have that this limit is- 1 /a 2 , as claimed. The above example was a rather quick and direct calculation; our second exam- ple, however, is a bit trickier: Lemma 3.2. The derivative of f ( x ) = e x at any point a is e a ....

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