Section 3.1/3.2: Rules of Differentiation

Transcription

1 : Rules of Differentiation Math February 2018

2 Overview 1 2

3 Four Theorem for Derivatives Suppose c is a constant an f, g are ifferentiable functions. Then x (c) = 0 x (x n ) = nx n 1 x [cf (x)] = c x [f (x)] x [f (x) ± g(x)] = x [f (x)] ± x [g(x)]

4 Fin the erivative of f (x) = x 2 x 1.

5 Fin the erivative of f (x) = x 2 x 1. Solution: We want x [f (x)]. Thus x [x 2 x 1] = x [x 2 ] x [x] [1] = 2x = 2x 1. x

6 Fin the erivative of f (x) = 5x 2 16x + 3 x + π.

7 Fin the erivative of f (x) = 5x 2 16x + 3 x + π. Solution: We want x [f (x)]. Thus x [5x 2 16x + 3 x + π] = x [5x 2 ] x [16x] + x [3 x] + x [π] = 5 x [x 2 ] 16 x [x] + 3 x [x 1 2 ] + 0 = 5(2x) 16(1) + 3( 1 2 x 1 2 ) = 10x x 1 2.

8 Fin the equation of the tangent line to the graph of f (x) = x + 1 x at the point (2, 5 2 ).

9 Fin the equation of the tangent line to the graph of f (x) = x + 1 x at the point (2, 5 2 ). Solution: First we nee to fin f (x). Using our rules of ifferentiation we get f (x) = 1 1. The slope of the tangent line x 2 at x = 1 will then be f (2) = 1 1 = = 3 4. Using point-slope form we get an equation for the tangent line at (2, 5 2 ) of y 5 2 = 3 (x 2). 4

10 When testing the brakes for a new race car, the istance s (in feet) travelle by the new car t secons after applying the brakes yiele a function s = 120t 15t 2. What is the velocity for the car at time t? What is the velocity when the brakes were first applie? How far i the car go before coming to a complete stop (i.e. when velocity was zero)?

11 It is a fact that the erivative of istance is velocity. Thus the velocity (in feet per secon) is given by the equation v = s = s t = t. The brakes were first applie at t = 0 so the velocity when the brakes were first applie as v = (0) = 120 feet per secon. To fin the stopping time we nee v = 0. Thus t = 0 implies that at t = 4 secons the car wasn t moving anymore. At this point the car ha travelle 120(4) 15(4) 2 = 240 feet.

12 Prouct an Quotient Rule Theorem Prouct an Quotient Rules If f an g are ifferentiable functions, then x [f (x)g(x)] = f (x)g (x) + g(x)f (x) an x [ f (x) g(x) ] = g(x)f (x) f (x)g (x) [g(x)] 2 Note that x [f (x)g(x)] f (x)g (x)!!!

13 Fin f (x) for f (x) = (x 2 x 1)(x 3 + 1). Solution: One way woul be to factor things out but let s just use the prouct rule. x [(x 2 x 1)(x 3 + 1)] = (x 2 x 1)(3x 2 ) + (x 3 + 1)(2x 1) = 3x 4 3x 3 3x 2 + 2x 4 + 2x x 3 1 = 5x 4 4x 3 3x 2 + 2x 1.

14 Fin h (x) for h(x) = (x 1)(x2 +x+1) x+1 Solution: For this problem we will nee to use the quotient an the prouct rule. To simplify things let s set f (x) = (x 1)(x 2 + x + 1) an g(x) = x + 1. Then f (x) = (x 1)(2x + 1) + (x 2 + x + 1) = 3x 2 an g (x) = 1. Now h(x) = f (x) g(x) so by the quotient rule we have x [ f (x) g(x) ] = g(x)f (x) f (x)g (x) [g(x)] 2 = (x + 1)(3x 2 ) (x 1)(x 2 + x + 1) (x + 1) 2

15 x Let f (x) =. Fin all points of the function where the tangent x 2 +1 line is horizontal. Solution: The tangent line being horizontal is the same as saying that f (x) = 0. Thus we nee to calculate f (x). f (x) = (x 2 + 1)(1) x(2x) (x 2 + 1) 2 = x (x 2 + 1) 2. Therefore we want 0 = 1 x2. Thus we want to know where our (x 2 +1) 2 numerator is equal to zero. In other wors 0 = 1 x 2 = (1 x)(1 + x). Thus our tangent line will be horizontal when x = ±1.