Calculus_Refresher - Calculus refresher Disclaimer I claim...

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Calculus refresher Disclaimer: I claim no original content on this document, which is mostly a summary-rewrite of what any standard college calculus book offers. (Here I’ve used Calculus by Dennis Zill.) I consider this as a brief refresher of the bare-bones calculus requirements we will be using during the course. No exercises are offered: any calculus book provides an insane amount of practice problems. 1. Rules of differentiation: Basics In what follows, let f ( x ) be a one-variable function, and denote its derivative by f 0 ( x ). The standard differentiation rules are the following: Theorem 1.1 (The Power Rule, 1) . Let n be a positive integer. Then d dx [ x n ] = nx n - 1 . (1) Example 1.2. The derivative of y = x 4 is given by dy dx = 4 x 4 - 1 = 4 x 3 . Theorem 1.3 (Derivative of a constant function) . If f ( x ) = k and k is a constant, then f 0 ( x ) = 0 . Theorem 1.4 (Derivative of a constant multiple of a function) . If c is any constant and f is a differentiable function, then d dx [ cf ( x )] = cf 0 ( x ) . (2) Example 1.5. Following theorems 1.1 and 1.4, the derivative of y = 3 x 5 is given by: dy dx = 3 · d dx x 5 = 3(5 x 4 ) = 15 x 4 . Theorem 1.6 (The Sum Rule) . Let f and g be two differentiable functions. Then d dx [ f ( x ) + g ( x )] = f 0 ( x ) + g 0 ( x ) . (3) Example 1.7. From theorems 1.1. and 1.6, the derivative of y = x 4 + x 3 is: dy dx = d dx x 4 + d dx x 3 = 4 x 3 + 3 x 2 . Theorem 1.8 (The Product Rule) . If f and g are differentiable functions, then d dx [ f ( x ) g ( x )] = f ( x ) g 0 ( x ) + g ( x ) f 0 ( x ) . (4) Example 1.9. The differential of y = ( x 3 - 2 x 2 + 4)(8 x 2 + 5 x ) is: dy dx = ( x 3 - 2 x 2 + 4) · d dx (8 x 2 + 5 x ) + (8 x 2 + 5 x ) · d dx ( x 3 - 2 x 2 + 4) = ( x 3 - 2 x 2 + 4)(16 x + 5) + (8 x 2 + 5 x )(3 x 2 - 4 x ) . Theorem 1.10 (The Quotient Rule) . If f and g are differentiable functions, then d dx " f ( x ) g ( x ) # = g ( x ) f 0 ( x ) - f ( x ) g 0 ( x ) [ g ( x )] 2 . (5) 1
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2 Example 1.11. The differential of y = 3 x 2 - 1 2 x 3 +5 x 2 +7 is: dy dx = (2 x 3 + 5 x 2 + 7) · d dx (3 x 2 - 1) - (3 x 2 - 1) · d dx (2 x 3 + 5 x 2 + 7) (2 x 3 + 5 x 2 + 7) 2 = (2 x 3 + 5 x 2 + 7) · (6 x ) - (3 x 2 - 1) · (6 x 2 + 10 x ) (2 x 3 + 5 x 2 + 7) 2 = - 6 x 4 + 6 x 2 + 52 x (2 x 3 + 5 x 2 + 7) 2 .
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