Derivatives

# Derivatives - h-P(3 h = lim h → 100 30(3 h 4(3 h 2(100 30...

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THE DERIVATIVE Whenever we calculate the slope of a tangent line, the velocity of an object, or any rates of change such as a rate of reaction in chemistry, a marginal cost in economics, or a population growth in biology, we encounter limits of a type lim x a f ( x ) - f ( a ) x - a = lim h 0 f ( a + h ) - f ( a ) h . Definition 1. The derivative of a function f at a number a is f 0 ( a ) = lim h 0 f ( a + h ) - f ( a ) h Note that we made a change of variable as x = a + h . Example 1. Find The derivative of the function f ( x ) = x 2 + x + 1 at 1. Solution f 0 (1) = lim h 0 f (1 + h ) - f (1) h = lim h 0 (1 + h ) 2 + (1 + h ) + 1 - 3 h = lim h 0 3 h + h 2 h = lim h 0 (3 + h ) = 3 . Example 2. A population of squirrels moves into a new region at time t = 0. At time t (in months) the population numbers P ( t ) = 100 + 30 t + 4 t 2 . (1) How long does it take for this population to double its initial size P (0)? (2) What is the rate of growth of the population when t = 3? Solution (1) We have to solve the quadratic equation 100 + 30 t + 4 t 2 = 2 P (0) = 200, and it gives us t = 2 . 5. (2) Rate of growth of the population when t = 3 is nothing more than P 0 (3). We can get P 0 (3) = lim h

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Unformatted text preview: h )-P (3) h = lim h → 100 + 30(3 + h ) + 4(3 + h ) 2-(100 + 30 × 3 + 4 × 3 2 ) h = lim h → 30 h + 24 h + 4 h 2 h = lim h → (54 + 4 h ) = 54 , thus the rate of growth is 54 / month. 1 2 THE DERIVATIVE Example 3. The cost (in dollars) of producing x units of a certain commodity is C ( x ) = 3000 + 5 x + 0 . 02 x 2 . (1) Find the average rate of change of C with respect to x when the production level is changed from x = 100 to x = 101. (2) Find the rate of change of the cost with respect to x when x = 100. Solution (1) Average rate of change is given by C (101)-C (100) 101-100 = 9 . 04 (2) Rate of change when x = 100 is given by P (100) = lim h → P (100 + h )-f (100) h = lim h → 3000 + 5(100 + h ) + 0 . 02(100 + h ) 2-(3000 + 5 × 100 + 0 . 02 × 100 2 ) h = lim h → 5 h + 4 h + 0 . 02 h 2 h = lim h → (9 + 0 . 02 h ) = 9...
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