multivariable_03_Rules_for_Finding_Derivatives_2up

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Unformatted text preview: 3 Rules for Finding Derivatives It is tedious to compute a limit every time we need to know the derivative of a function. Fortunately, we can develop a small collection of examples and rules that allow us to compute the derivative of almost any function we are likely to encounter. Many functions involve quantities raised to a constant power, such as polynomials and more complicated combinations like y = (sin x ) 4 . So we start by examining powers of a single variable; this gives us a building block for more complicated examples. h eÓÛ eÖÙ Ðe We start with the derivative of a power function, f ( x ) = x n . Here n is a number of any kind: integer, rational, positive, negative, even irrational, as in x π . We have already computed some simple examples, so the formula should not be a complete surprise: d dx x n = nx n − 1 . It is not easy to show this is true for any n . We will do some of the easier cases now, and discuss the rest later. The easiest, and most common, is the case that n is a positive integer. To compute the derivative we need to compute the following limit: d dx x n = lim Δ x → ( x + Δ x ) n − x n Δ x . For a specific, fairly small value of n , we could do this by straightforward algebra. 43 44 Chapter 3 Rules for Finding Derivatives EXAMPLE 3.1 Find the derivative of f ( x ) = x 3 . d dx x 3 = lim Δ x → ( x + Δ x ) 3 − x 3 Δ x . = lim Δ x → x 3 + 3 x 2 Δ x + 3 x Δ x 2 + Δ x 3 − x 3 Δ x . = lim Δ x → 3 x 2 Δ x + 3 x Δ x 2 + Δ x 3 Δ x . = lim Δ x → 3 x 2 + 3 x Δ x + Δ x 2 = 3 x 2 . The general case is really not much harder as long as we don’t try to do too much. The key is understanding what happens when ( x + Δ x ) n is multiplied out: ( x + Δ x ) n = x n + nx n − 1 Δ x + a 2 x n − 2 Δ x 2 + ··· + + a n − 1 x Δ x n − 1 + Δ x n . We know that multiplying out will give a large number of terms all of the form x i Δ x j , and in fact that i + j = n in every term. One way to see this is to understand that one method for mulitplying out ( x + Δ x ) n is the following: In every ( x + Δ x ) factor, pick either the x or the Δ x , then multiply the n choices together; do this in all possible ways. For example, for ( x + Δ x ) 3 , there are eight possible ways to do this: ( x + Δ x )( x + Δ x )( x + Δ x ) = xxx + xx Δ x + x Δ xx + x Δ x Δ x + Δ xxx + Δ xx Δ x + Δ x Δ xx + Δ x Δ x Δ x = x 3 + x 2 Δ x + x 2 Δ x + x Δ x 2 + x 2 Δ x + x Δ x 2 + x Δ x 2 + Δ x 3 = x 3 + 3 x 2 Δ x + 3 x Δ x 2 + Δ x 3 No matter what n is, there are n ways to pick Δ x in one factor and x in the remaining n − 1 factors; this means one term is nx n − 1 Δ x . The other coefficients are somewhat harder to understand, but we don’t really need them, so in the formula above they have simply been called a 2 , a 3 , and so on. We know that every one of these terms contains Δ x to at least the power 2. Now let’s look at the limit: d dx x n = lim Δ x → ( x + Δ x ) n − x n...
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## This note was uploaded on 12/01/2011 for the course MATH 305 taught by Professor Guichard during the Fall '11 term at Whitman.

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