This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
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. he e e 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 . 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 x ( x + x ) n x n 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 x ( x + x ) 3 x 3 x . = lim x x 3 + 3 x 2 x + 3 x x 2 + x 3 x 3 x . = lim x 3 x 2 x + 3 x x 2 + x 3 x . = lim x 3 x 2 + 3 x x + x 2 = 3 x 2 . The general case is really not much harder as long as we dont try to do too much. The key is understanding what happens when ( x + x ) n is multiplied out: ( x + x ) n = x n + nx n 1 x + a 2 x n 2 x 2 + + + a n 1 x x n 1 + x n . We know that multiplying out will give a large number of terms all of the form x i 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 + x ) n is the following: In every ( x + x ) factor, pick either the x or the x , then multiply the n choices together; do this in all possible ways. For example, for ( x + x ) 3 , there are eight possible ways to do this: ( x + x )( x + x )( x + x ) = xxx + xx x + x xx + x x x + xxx + xx x + x xx + x x x = x 3 + x 2 x + x 2 x + x x 2 + x 2 x + x x 2 + x x 2 + x 3 = x 3 + 3 x 2 x + 3 x x 2 + x 3 No matter what n is, there are n ways to pick x in one factor and x in the remaining n 1 factors; this means one term is nx n 1 x . The other coefficients are somewhat harder to understand, but we dont 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 x to at least the power 2. Now lets look at the limit: d dx x n = lim x ( x + x ) n x n...
View Full Document
- Fall '11