CHAPTER 8: DIFFERENTIAL CALCULUS

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1 CHAPTER 8: DIFFERENTIAL CALCULUS 1. Rules of Differentiation As we ave seen, calculating erivatives from first principles can be laborious an ifficult even for some relatively simple functions. It is clearly not a very feasible approac if we want to fin te erivative of a more complicate function suc as 3 x + 11x 4 f(x) = 2 x 4x. 7 However, te power of calculus as a tool lies in te fact tat functions suc as tis can be ifferentiate witout great effort an (after some practice) wit almost no tougt. Just as for limits, tere are rules for calculating te erivatives of more complicate function wic are constructe from simple ingreients (suc as power functions). It is te use of tese rules to ifferentiate functions wic is calle te Differential Calculus Te Power Rule for Differentiation. We ave alreay calculate (from first principles) te erivatives of several power functions. Example 1.1. Te function f(x) = x 0 is just te constant function f(x) = 1. Tus, x (x0 ) = 0. Example 1.2. Te function f(x) = x 1 is just te ientity function f(x) = x. Tus, x (x1 ) = 1. Example 1.3. We ave calculate te erivative of f(x) = x 2 : x (x2 ) = 2x. Example 1.4. Te functions f(x) = 1/x is te power function f(x) = x 1. Tus x (x 1 ) = x 2. Example 1.5. Te function f(x) = x is te power function f(x) = x 1/2. Tus, x (x1/2 ) = 1 2 x 1/2. 1

2 2 First Science MATH1200 Calculus Observe tat, in eac case, wen we ifferentiate a power of x, te resulting power is te original power reuce by 1, an tat tis new power of x is multiplie by te original exponent. Tis is te general rule wic works for every function: For any r R, x xr = rx r 1. i.e., if f(x) = x r, ten f (x) = rx r 1. Of course, tere is noting privilege about te letter x ere: Example 1.6. Example 1.7. t tr = rt r 1 s sr = rs r 1 w wr = rw r 1 η ηr = rη r 1 Tom Tomr = rtom r 1... etc x (x3 ) = 3x 2. x (x7 ) = 7x 6. Example 1.8. Tis general power rule is consistent wit te calculations tat we ave alreay mae for special power functions, even in te case r = 0 an r = 1: Example 1.9. x (x0 ) = 0x 1 = 0 x = 1 1 x0 = 1 1 = 1 x xπ = πx π 1 Some care nees to be taken wit fractional an negative powers. Example x (1/x4 ) = x (x 4 ) = 4x 5 = 4 x 5

3 First Science MATH1200 Calculus 3 Example x (1/ 3 x) = x (x 1/3 ) = 1 3 x 4/3 = x 4 Example Calculate te erivative of te function f(x) = x 3 / 3 x 2. Solution: From te properties of exponents we ave Tus x 3 3 x 2 = x3/2 x = x 3 2/ = x 5/6. f (x) = x x5/6 = 5 6 x 1/6 = x Te Sum Rule / Difference Rule for Differentiation. If we know ow to ifferentiate two functions f(x) an g(x), te sum rule tells us ow to ifferentiate te sum of te two functions. {f(x) ± g(x)} = x x f(x) ± x g(x) Example Example Example x (x3 + x) = x (x3 ) + x ( x) = 3x x. x ( x x x ) = 1 2 x 5x4 1 x. 2 x (3 + x2 + 3 x) = 0 + 2x x 2/3 = 2x x Te Constant Rule for Differentiation. x C f(x) = C x f(x) (C a constant) Example x (10 5 x) = 10 x 5 x = x 4/5.

4 4 First Science MATH1200 Calculus Example x (15x7 ) = 15 7x 6 = 105x 6. We can combine te constant an sum/ifference rules: Example x (5x3 6x 2 + 4x + 12) = 5 3x 2 6 2x = 15x 2 12x + 4 Te sum rule for erivatives resembles te sum rule for limits. Tis is because te former is a straigtforwar consequence of te latter: Proof of te sum rule: Let s(x) = f(x) + g(x) Ten x s(x) = s (x) 0 s(x + ) s(x) [f(x + ) + g(x + )] [f(x) + g(x)] 0 { } f(x + ) f(x) g(x + ) g(x) + 0 f(x + ) f(x) g(x + ) g(x) + lim 0 0 (sum rule for limits) = f (x) + g (x) 1.4. Te Prouct Rule for Differentiation. If u an v are functions of x ten (uv) = uv x x + v u x i.e., (uv) (x) = u(x) v (x) + v(x) u (x) Example x ( }{{} x5 ( x + x 3 )) } {{ } u v = x 5 x ( x + x 3 ) + ( x + x 3 ) x x5 = x 5 1 ( 2 x + 3x2 ) + ( x + x 3 ) 5x 4 [Of course, we coul also procee as follows: x 5 ( x + x 3 ) = x 5 (x 1/2 + x 3 ) = x 5 x 1/2 + x 5 x 3 = x 11/2 + x 8 an now use te sum an power rules. We will see more interesting uses of te prouct rule wen we learn to ifferentiate

5 First Science MATH1200 Calculus 5 some more interesting functions: trigonometric functions, exponentials an logaritms.] Example Suppose tat we ave a function f(x) an tat all we know about it is its value an its rate of growt at x = 3. Suppose, for example, tat we know tat f(3) = 15 an f (3) = 11. Let g(x) be te function g(x) = x 2 f(x). Wat is g (3)? Solution: By te prouct rule, Tus g (x) = x 2 f (x) + 2xf(x). g (3) = 9f (3) + 6f(3) = ( 15) = = 9. [Te prouct rule for limits says tat te limit of a prouct is just te prouct of te two limits. Wy, ten, is te prouct rule for ifferentiation suc a complicate rule? Proof of te Prouct Rule: Let p(x) = u(x) v(x) (uv) = x p p(x+) p(x) (x) 0 0 u(x+) v(x+) u(x) v(x) 0 u(x+)v(x+) u(x)v(x+)+u(x)v(x+) u(x)v(x) 0 u(x+) u(x) 0 [u(x+) u(x)] v(x+)+u(x) [v(x+) v(x)] lim 0 v(x + ) + u(x) lim 0 v(x+) v(x) = u (x) v(x) + u(x) v (x).] 1.5. Te Quotient Rule for Differentiation. If u an v are two functions of x ten ( u ) = v u u v x x x v v 2 Equivalently, letting u = f(x), v = g(x), we can write tis rule in te form: ( ) f(x) = g(x)f (x) f(x)g (x) x g(x) g(x) 2 Caution: Because of te minus sign in te formula, te orer in wic te terms appear is very significant. Te formula can be expresse as bottom erivative of top top erivative of bottom (bottom) 2.

6 6 First Science MATH1200 Calculus Example x ( 3x + 2 x 3 + x ) = (x3 + x) (3x + 2) (3x + 2) x x (x3 + x) (x 3 + x) 2 = (x3 + x) 3 (3x + 2)(3x 2 + 1) (x 3 + x) 2 = (3x3 + 3x) (9x 3 + 3x + 6x 2 + 2) (x 3 + x) 2 = 6x3 6x 2 2 (x 3 + x) 2. Example Fin te equation of te tangent line to te grap of te function f(x) = x3 + 1 x 2 + x 1 at x = 1 (y = 2). Solution: We first fin te erivative of f(x) at an arbitrary point x: f (x) = x ( x x 2 + x 1 ) = (x2 + x 1) x (x3 + 1) (x 3 + 1) x (x2 + x 1) = (x2 + x 1)3x 2 (x 3 + 1)(2x + 1) = (3x4 + 3x 3 3x 2 ) (2x 4 + x 3 + 2x + 1) = x4 + 2x 3 3x 2 2x 1 Tus f (1) = = Tis is te slope of te tangent line. Te tangent line contains te point (1, 2). Terefore, te equation of te line is y 2 = 3(x 1) = 3x + 3. So y = 3x + 5 is te line in question.

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