Section 3.3. Differentiation of Polynomials and Rational Functions. Difference Equations to Differential Equations

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1 Difference Equations to Differential Equations Section 3.3 Differentiation of Polynomials an Rational Functions In tis section we begin te task of iscovering rules for ifferentiating various classes of functions. By te en of Section 3.5 we will be able to ifferentiate any algebraic or trigonometric function as a matter of routine witout reference to te limits use in Section 3.2. Differentiation of polynomials We first note tat if f is a first egree polynomial, say, f(x) = ax + b for some constants a an b, ten f is an affine function an ence its own best affine approximation. Tus f (x) = a for all x. In particular, if f is a constant function, say, f(x) = b for all x, ten f (x) = 0 for all x. Next we consier te case of a monomial f(x) = x n, were n is a positive integer greater tan 1. Ten Now f (x) 0 f(x + ) f(x) (x + ) n x n. (3.3.1) 0 (x + ) n = x n + nx n 1 + R(), (3.3.2) were R() represents te remaining terms in te expansion. Since every term in R() as a factor of raise to a power greater tan or equal to 2, it follows tat R() is o(). Hence we ave f x n + nx n 1 + R() x n (x) 0 nx n 1 + R() 0 ( nx n 1 + R() ) 0 = nx n 1 R() + lim 0 = nx n 1. Since from our previous result f (x) = 1 wen f(x) = x, tis formula also works in te case n = 1. Hence we ave te following proposition. Proposition For any positive integer n, x xn = nx n 1. (3.3.3) 1 Copyrigt c by Dan Slougter 2000

2 2 Differentiation of Polynomials an Rational Functions Section 3.3 If f(x) = x 3, ten f (x) = 3x 2, as we saw in an example in Section 3.2. Similarly, t t5 = 5t 4. Hence, for example, te equation of te line tangent to te curve x = t 5 at ( 1, 1) is or x = 5(t + 1) 1, x = 5t + 4. Once we establis results for te erivative of a constant times a function an for te erivative of te sum of two functions, similar to te results we ave for limits, we will be able to easily ifferentiate any polynomial. So suppose f is a ifferentiable function an let k(x) = cf(x), were c is any constant. Ten k (x) 0 k(x + ) k(x) 0 cf(x + ) cf(x) 0 c(f(x + ) f(x)) = c lim 0 f(x + ) f(x) = cf (x). Tat is, te erivative of a constant times a function is te constant times te erivative of te function. Proposition If f is ifferentiable an c is any constant, ten x (cf(x)) = c f(x). (3.3.4) x If f(x) = 14x 3, ten f (x) = (14)(3x 2 ) = 42x 2. Now suppose f an g are bot ifferentiable functions an let k(x) = f(x) +. Ten k k(x + ) k(x) (x) 0 (f(x + ) + g(x + )) (f(x) + ) 0 0 ( f(x + ) f(x) + f(x + ) f(x) 0 = f (x) + g (x). g(x + ) ) + lim 0 g(x + )

3 Section 3.3 Differentiation of Polynomials an Rational Functions 3 Hence te erivative of te sum of two functions is te sum of teir erivatives. A similar argument woul sow tat te erivative of te ifference of two functions is te ifference of teir erivatives. Proposition If f an g are bot ifferentiable, ten (f(x) + ) = x x f(x) + (3.3.5) x an (f(x) ) = x x f(x). (3.3.6) x Putting te preceing results togeter, we are now in a position to easily ifferentiate any polynomial, as te next examples will illustrate. Suppose f(x) = 3x 5 6x 2 + 2x 16. Ten f (x) = x (3x5 6x 2 + 2x 16) = x (3x5 ) x (6x2 ) + x (2x) x (16) = 3 x x5 6 x x2 + 2 x x 0 = (3)(5x 4 ) (6)(2x) + (2)(1) = 15x 4 12x + 2. Of course, it is not necessary to write out in etail all te steps in ifferentiating a polynomial as we i in te preceing example. For example, if g(t) = 3t 12 6t 2 + t, ten g (t) = (3)(12t 11 ) (6)(2t) + 1 = 36t 11 12t + 1. In particular, since g(1) = 2 an g (1) = 25, te best affine approximation to g at t = 1 is T (t) = 25(t 1) 2 = 25t 27. Differentiation of rational functions We next consier te problem of ifferentiating te quotient of two functions wose erivatives are alreay known. In particular, combining tis result wit our result for polynomials will enable us to easily ifferentiate any rational function. We migt ope tat, analogous to te last two results an te relate results for limits, te erivative of te quotient of two functions woul be equal to te quotient of teir erivatives. Tis turns out not to be true; neverteless, tere is a nice rule for ifferentiating quotients.

4 4 Differentiation of Polynomials an Rational Functions Section 3.3 Suppose f an g are bot ifferentiable functions an let k(x) = f(x). Ten, at all points were 0, k k(x + ) k(x) (x) f(x + ) g(x + ) f(x) f(x + ) g(x + )f(x) g(x + ) f(x + ) g(x + )f(x) 0 g(x + ) It turns out tat by aing an subtracting te term f(x) (a stanar matematical trick of aing 0) in te numerator, we can simplify tis limit into a form tat we can evaluate. Tat is, Now an k (x) 0 f(x + ) g(x + )f(x) g(x + ) f(x + ) f(x) + f(x) g(x + )f(x) 0 g(x + ) (f(x + ) f(x)) f(x)(g(x + ) ) 0 g(x + ) f(x + ) f(x) g(x + ) f(x). 0 g(x + ) f(x + ) f(x) lim 0 g(x + ) lim f(x) 0 = lim 0 f(x + ) f(x) = f(x) lim 0 g(x + ). = f (x), (3.3.7) = f(x)g (x), (3.3.8) lim g(x + ) = lim g(x + ) = = 0 0 ()2, (3.3.9) were te limits in (3.3.7) an (3.3.8) follow from te ifferentiability of f an g, wile te limit in (3.3.9) follows from te continuity of g (wic is a consequence of te ifferentiability of g). Putting everyting togeter, we ave a result known as te quotient rule. k (x) = f (x) f(x)g (x) () 2, (3.3.10)

5 Section 3.3 Differentiation of Polynomials an Rational Functions 5 Quotient Rule If f an g are bot ifferentiable, ten x at all points were 0. f(x) = Suppose f(x) = 2x + 1 x 2. Ten f(x) f(x) x x () 2 (3.3.11) (x 2) (2x + 1) (2x + 1) (x 2) f (x) = x x (x 2) 2 (x 2)(2) (2x + 1)(1) = (x 2) 2 2x 4 2x 1 = (x 2) 2 5 = (x 2) 2. Hence, for example, f(3) = 7 an f (3) = 5, so te equation of te line tangent to te grap of f at (3, 7) is y = 5(x 3) + 7, or y = 5x Suppose g(z) = 1 z 2. Ten g (z) = z 2 z (1) (1) z (z2 ) z 4 = (z2 )(0) 2z z 4 = 2 z 3. Note tat we may write tis result in te form wic is consistent wit our previous result z z 2 = 2z 3, z zn = nz n 1. However, we erive te latter uner te assumption tat n was a positive integer. We will now sow tat we can exten tis result to te case of negative integer exponents.

6 6 Differentiation of Polynomials an Rational Functions Section 3.3 Suppose f(x) = x n, were n is a negative integer. Ten, using te quotient rule an te fact tat n > 0, f (x) = x xn = 1 x x n x n (1) (1) = x x (x n ) x 2n = (x n )(0) ( nx n 1 ) x 2n = nx n 1 x 2n = nx n 1+2n = nx n 1. We can now state te more general result. Proposition For any integer n 0, x xn = nx n 1. (3.3.12) If f(x) = 1 x, ten f (x) = x x 1 = x 2 = 1 x 2. Similarly, 5 x x 3 = x (5x 3 ) = 15x 4 = 15 x 4. We will eventually see tat (3.3.12) ols for rational an irrational exponents as well. We will consier te rational case in Section 3.4, but we will not ave te tools for anling te irrational case until we iscuss exponential an logaritm functions in Capter 6. Differentiation of proucts We will close tis section wit a iscussion of a rule for ifferentiating te prouct of two functions. Since te prouct of two rational functions is again a rational function, tis will not exten te class of functions tat we know ow to ifferentiate routinely. However, tis rule will be very useful in te future an, even at te present point, can elp simplify some problems.

7 Section 3.3 Differentiation of Polynomials an Rational Functions 7 Suppose f an g are bot ifferentiable an k(x) = f(x). Ten k (x) 0 k(x + ) k(x) f(x + )g(x + ) f(x). (3.3.13) 0 Aing an subtracting f(x + ) in te numerator (again, te matematical trick of aing 0 in a useful manner) will elp simplify tis limit. Namely, k f(x + )g(x + ) f(x + ) + f(x + ) f(x) (x) 0 f(x + )(g(x + ) ) + (f(x + ) f(x)) 0 0 ( f(x + ) Now g(x + ) lim f(x+) 0 ( g(x + ) f(x + ) f(x) + 0 f(x+) lim 0 g(x + ) )). = f(x)g (x) (3.3.14) an f(x + ) f(x) lim 0 = lim 0 f(x + ) f(x) = f (x), (3.3.15) were, as wit te erivation of te quotient rule, we ave use te ifferentiability of f an g as well as te continuity of f in evaluating te limits. Putting everyting togeter, we ave k (x) = f(x)g (x) + f (x). (3.3.16) a result known as te prouct rule. Prouct Rule ten If If f an g are bot ifferentiable, ten f(x) = f(x) + f(x). (3.3.17) x x x f(x) = (x 4 3x 2 + 6x 3)(6x 3 + 2x + 5), f (x) = (x 4 3x 2 + 6x 3) x (6x3 + 2x + 5) + (6x 3 + 2x + 5) x (x4 3x 2 + 6x 3) = (x 4 3x 2 + 6x 3)(18x 2 + 2) + (6x 3 + 2x + 5)(4x 3 6x + 6). Of course, in tis example, f is just a polynomial so we coul also fin f by multiplying out te two factors of f an ifferentiating te polynomial term by term as usual. However,

8 8 Differentiation of Polynomials an Rational Functions Section 3.3 te prouct rule gives us a quicker route to te erivative. Altoug te result is not simplifie into te stanar form of a polynomial, for most applications tis form is just as useful as any oter. It is wort noting tat altoug we can now ifferentiate any rational function in teory, in practice our metos may not be te most useful. For example, te function f(x) = (x 2 + 1) 567 is a polynomial, an so we know ow to ifferentiate it. However, at tis point te only way we coul perform te ifferentiation woul be to expan f(x) into stanar polynomial form an ten ifferentiate term by term. In Section 3.4 we will learn ow to anle tis problem more irectly. At te same time we will exten te class of functions tat we can ifferentiate routinely to inclue all algebraic functions. Problems 1. Fin te erivative of eac of te following functions. (a) f(x) = x 3 + 6x (b) = 13x 5 6x (c) g(t) = 3t 6t 2 () y(t) = 4t 3 18t + 3 (e) f(t) = (3t 6) 2 (f) f(x) = (4x + 5)(6x 2 1) 2. Fin te erivative of eac of te following functions. (a) f(x) = (2x + 1) 2 (b) g(t) = (t 2 3) 3 (c) = x 3 2x + 5 (e) f(t) = 3t4 8t + 1 2t (g) (t) = 3 t (i) (z) = 8z 3 1 2z () (s) = 2s s2 s (f) x(t) = 3 t 3 16t2 () f(x) = 41 3x 7 (j) f(s) = 31 s s 2 16s 3. For eac of te following, make use of te prouct rule in fining te erivative of te epenent variable wit respect to te inepenent variable. (a) s = (t 2 6t + 3)(8t 4 + 6t 2 7) (b) q = (13t 4 + 5t)(3t 5 + 4t t 31) (c) y = (x 2 2x + 3)(2x x 6)(3x 2 4x + 1) () z = (x2 3x + 6)(8x 2 + 3x 2) x 2 6

9 Section 3.3 Differentiation of Polynomials an Rational Functions 9 4. Suppose f(2) = 2, f (2) = 6, g(2) = 3, an g (2) = 4. Fin k (2) for eac of te following. (a) k(x) = f(x) (b) k(x) = f(x) (c) k(x) = f(x)() 2 f(x) f(x) () k(x) = 5. Suppose an object moves along te x-axis so tat its position at time t is x = t + t3 6. (a) Fin te velocity, v(t) = ẋ(t), of te object. (b) Wat is v(0)? Wat oes tis say about te irection of motion of te object at time t = 0? (c) Wen is te object at te origin? Wat is te velocity of te object wen it is at te origin? () For wat values of t is te object moving towar te rigt? (e) For wat values of t is te object moving towar te left? (f) Wat is appening at te points were v(t) = 0? (g) Fin te acceleration of te object, a(t) = v(t). () Wen is te acceleration positive? Wen is it negative? (i) Notice tat v(1) < 0 an a(1) > 0. Wat oes tis say about te motion at time t = 1? 6. (a) Using only te prouct rule an te fact tat x = 1, sow tat x x x2 = 2x. (b) Now use te prouct rule to sow tat x x3 = 3x 2. (c) Let n > 1 an suppose we know tat x xm = mx m 1 for all m < n. Use te prouct rule to sow tat x xn = nx n 1.

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