Section 3.1: Derivatives of Polynomials and Exponential Functions

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1 Section 3.1: Derivatives of Polynomials and Exponential Functions In previous sections we developed te concept of te derivative and derivative function. Te only issue wit our definition owever is tat it requires taking te it of a difference quotient. We ave seen tat even for easy functions, tis can be difficult. In te next few sections, we develop some rules wic will allow us to differentiate some of te more common functions. We start wit te natural exponential function and polynomials. 1. Derivatives of Power Functions Te easiest type of functions to differentiate are power functions. To obtain a rule for power functions, we start wit te easiest - a constant function (or zero power). Result 1.1. If f(x) = c, ten f (x) = 0. Proof. f (x) f(x + ) f(x) Next we consider power functions. c c 0 = 0. Result 1.2. (Te Power Rule) Suppose tat f(x) = x n were n is any number. Ten f (x) = nx n 1. Proof. Te proof for general values of n is fairly complicated, so we sall just consider te proof for a couple of positive integral values of n (te ideas for iger powers is similar). Wen n = 1, we ave f(x) = x. Tus we ave f f(x + ) f(x) (x) Wen n = 2, we ave f (x) f(x + ) f(x) x + x (x + ) 2 x 2 2x + 2 2x = 2 x 1. Oter values of n follow a similar argument. We illustrate wit some examples. = = 1 x0. x 2 + 2x + 2 x 2 Example 1.3. Use te power rule to differentiate te following: (i) f(x) = 2 We ave f(x) = 2 = 2x 0, so f (x) = 0 2x 1 = 0. 1

2 2 (ii) f(x) = x πe Using te power rule, we ave f (x) = π e x πe 1. (iii) f(x) = x 3 (iv) x 2/5 Converting into a power, we ave f(x) = x 3 = x 3/2, so f (x) = 3x1/2 2. Using te power rule, we ave f (x) = 5x 7/2. 2 Example 1.4. Find te equation of te tangent line to y = x 2 at x = 2. First, we use te power rule to find te derivative of f(x). f (x) = 2x 2 1. Te derivative of f(x) at x = 2 gives us te slope of te tangent line at x = 2, so f (2) = Using point sloope form, we ave y = (x 2) Derivatives of Polynomials We would like to transfer our knowledge of power functions to polynomials and oter similar functions. However, in order to do tis, we need to examine te linear properties of te derivative. Result 2.1. (Linearity of te Derivative) Suppose f(x) and g(x) are differentiable and c is a constant. Ten we ave te following: (i) (Constant multiple) If (x) = cf(x), ten (x) = cf (x). (ii) (Sum Rule) If (x) = f(x) + g(x), ten (x) = f (x) + g (x). (iii) (Difference Rule) If (x) = f(x) g(x), ten (x) = f (x) g (x). We illustrate wy tese are true by proving te sum rule. Proof. (x) (x + ) (x) f(x + ) + g(x + ) (f(x) + g(x)) (f(x + ) f(x)) + (g(x + ) g(x)) (f(x + ) f(x)) (g(x + ) g(x)) + = f (x) + g (x)

3 Observe tat tis result now allows us to differentiate any polynomial (and in fact many different algebraic functions). We illustrate wit some examples. Example 2.2. (i) If f(x) = 3x 2 + 2x 1, find f (x) We ave f (x) = 6x + 2 using te power rule, te sum rule, te difference rule and te constant multiple rule. (ii) If f(x) = x 3 3x 2 + 3x, determine were te tangent line is orizontal and write down te equations for tangent lines at tese points. We ave f (x) = 3x 2 6x+3. Te tangent line is orizontal wen f (x) = 0, so 3x 2 6x + 3 = 0. Factoring, we ave 3(x 2 2x + 1) = 3(x 1) 2 = 0, so te only place f (x) = 0 is wen x = 1. Using point slope form, we get y = 0(x 1) + 1 = Derivatives of Exponential Functions Te last type of basic function we sall consider is te exponential function. In tis section we sall only be able to derive a formula for te derivative of te natural exponential function f(x) = e x - for oter exponential functions, we sall come up wit a formula wic will allow us to determine te derivative in a future section. We start wit te general exponential function. Result 3.1. Suppose f(x) = a x. Ten f (x) = Ca x were C is te constant a 1 C. Proof. Tis question was completed for a omework assignment. For completion, we sall do it again. We ave f a x+ a x (x) were a x (a 1) = a x (a 1) a 1 C. Tis means tat if we can evaluate te it = Ca x C a 1 ten we can determine te derivative of te general exponential function f(x) = a x. However, tis is no easy problem and we will ave to return to it in a later section. Wit te knowledge we currently ave toug, we can evaluate tis it for te function f(x) = e x. 3

4 4 Result 3.2. e 1 = 1. Proof. To evaluate tis it, we need to return to te definition of te number e. Recall tat ( ) 1 e 1 +. Tis means or and so e 1 + (e 1) e 1 = 1. Putting tese two previous results togeter, we get te derivative of te natural exponential function. Specifically, we ave: Result 3.3. (Derivative of te Natural Exponential Function) If f(x) = e x, ten f (x) = e x. We finis wit some examples. Example 3.4. Find te derivatives of te following functions. (i) f(x) = x + e x Using te sum law, te power law and te exponential law, we ave f (x) = 1 + e x. (ii) g(x) = e x + x 2 Using te sum law, te power law and te exponential law, we ave f (x) = 2x + e x. (iii) (x) = e x + x e Using te sum law, te power law and te exponential law, we ave f (x) = e x + ex e 1. Example 3.5. (i) Write down a general formula for te equation of te tangent line to f(x) = e x at x = a. First, te derivative of f(x) = e x is f (x) = e x. Terefore, te slope at x = a is f (a) = e a. Terefore, using point-slope form, te equation of te tangent line to f(x) = e x at (a, e a ) will be y = e a (x a) + e a..

5 (ii) Use your answer to find te intercept of te equation of a tangent line to f(x) = e x at x = a. From te previous question, we know te equation of te tangent line. Terefore, by setting tis equal to 0, we can determine te x-value of te intercetpt: 0 = e a (x a) + e a, so (x a)+1 = 0 or x = 1 a. Tus te intercept is at te point (1 a, 0). 5