More calculations of the derivative

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1 1 More calculations of the derivative There are more quick rules to help differentiate functions easily. Theorem (Product Rule) Dx [ f(x) g(x) ] = f(x) Dx[g(x)] + g(x) Dx[f(x)]. Using the prime notation, this says [f(x) g(x)] ' = f(x) g'(x) + g(x) f '(x). I think of this as saying: The derivative of a product is the first function times the derivative of the second function plus the second function times the derivative of the first. I don't memorize the formula using f and g, since when we use it the functions may have different names. A very common error is to hope that [f(x) g(x)] ' = f ' (x) g'(x). Nevertheless, this hope is not realized. Example. Let f(x) = (x x ) (2 x x 2 + 5) (a) Find f '(x). (b) Find f ' (1). (c) Find the slope of the tangent line where x = 1. (a) f '(x) = (x x ) (2 x x 2 + 5)' + (2 x x 2 + 5)(x x )' = (x x ) (6 x x ) + (2 x x 2 + 5)(30 x x 11 ) Multiplying out does not help, so you don't need to do it. (b) f ' (1) = (1+2+4)(6+8) + (2+4+5) (30+24) = (7) (14) + (11)(54) = 692 (c) The slope is f ' (1) = 692. Example. Let's do one we can check easily. If f(x) = (x 3 x 5 ) we find f ' (x) in two ways: (i) By the product rule, we get x 3 (5x 4 ) + x 5 (3x 2 ) = 8 x 7 (ii) Directly, f(x) = x 8, so f ' (x) = 8 x 7 So the two methods agree. Theorem. (Quotient Rule) Dx [ f(x) / g(x) ] = g(x) Dx[f(x)] - f(x) Dx[g(x)]. [g(x)] 2 Alternatively, [ f(x) / g(x) ] ' = g(x) f '(x) - f(x) g '(x) [g(x)] 2 Think of this as saying denominator * derivative of numerator - numerator * derivative of denominator

2 2 all divided by the denominator squared. A common error is to say that the derivative should be f'(x) / g'(x); but this is false. Example. Find the derivative of (3x+4) (2x-1) The derivative is (2x-1) Dx(3x+4) - (3x+4) Dx(2x-1) (2x-1) 2 = (2x-1) (3) - (3x+4) (2) (2x-1) 2 = 6x-3-6x-8 (2x-1) 2 = - 11 (2x-1) 2 Example. If f(x) = 2 x x 17 4x 3 + x find f '(x). f ' (x) = (4x 3 + x 2 + 1) Dx(2 x x 17 ) - (2 x x 17 )Dx(4x 3 + x 2 + 1) (4x 3 + x 2 + 1) 2 = (4x 3 + x 2 + 1) (40 x x 16 ) - (2 x x 17 )(12x 2 + 2x ) (4x 3 + x 2 + 1) 2 Multiplying out does not help, so you don't need to simplify. Here is another use of the quotient rule: Example. Find the derivative of 4 x 7 The derivative is x 7 Dx(4) - 4 Dx(x 7 ) (x 7 ) 2 = x 7 (0)- 4 (7x 6 ) x 14 = - 28x 6 x 14

3 3 = - 28 x 8 Theorem. If f(x) = x k, where k is a negative integer then f ' (x) = k x k-1. (So the formula is the same as when k is a positive integer.) Example. Find the derivative of g(x) = x -4. g'(x) = -4 x -5 Example. Find the derivative of g(x) = 3 / x 5 Solution. g(x) = 3 x -5 Hence g'(x) = 3 (-5) x -5-1 = -15 x -6 = -15 / x 6 Example. Let f(x) = 4 x 3-7/x 2 + 5/x + 3. (a) Find f '(x) (b) Find the equation of the line tangent to the curve y = f(x) at x = 1. f(x) = 4 x 3-7 x x Hence f ' (x) = 12x x -3-5 x -2 (b) Note f(1) = = 5 f '(1) = = 21 Hence y = (x-1) is the tangent line y = x - 21 y = 21 x - 16 Example. Let y = 2 t 3-3 t t /t 2-3/t (a) Find dy/dt (b) Find the instantaneous rate of change when t = 2. (a) dy/dt just means the derivative of y (when t is the variable). Reexpress y using negative exponents: y = 2 t 3-3 t t t -2-3t -1 Now dy/dt = 6 t 2-6 t t t -2 (b) The instantaneous rate is just the derivative. When t = 2, dy/dt = 6 (2) 2-6 (2) (2) (2) -2 = Example. Find the derivative of f(x)= 5 x x x 3 in two different ways. (a) f(x) = 5 x + 7 x x -3 So f '(x) = 5-7 x -2-6 x -4 (b) Or use the quotient rule: f ' (x) = x 3 (20 x x) - (5 x x 2 +2)(3x 2 ) (x 3 ) 2

4 4 = 20 x x 4 - (15 x x 4 +6x 2 ) (x 6 ) = 20 x x 4-15 x 6-21 x 4-6x 2 x 6 = 5 x 6-7 x 4-6x 2 x 6 = 5 x 6-7 x 4-6x 2 x 6 This form of the answer is acceptable. The Chain Rule Composition of functions. Often two functions are combined in another way. The composition g f of the functions f and g is the function (g f)(x) = g(f(x)). This means, "do the function f to x, then do g to the result." Example. g(x) = x 2 and f(x) = (3x+1). Then (g f)(x) = g(f(x)) = g(3x+1) = (3x+1) 2. But (f g)(x) = f(g(x)) = f(x 2 ) = 3x Usually g f and f g are not the same. Example. Write (2x+3) 4 as g f. In g f, g tells the last thing done. So g(x) = x 4. Simultaneously, f tells what g is done to, so f(x) = 2x+3. Example. Write (5x 3 + 7) as g f. g(x) = x since the last thing done is to take the square root. And f(x) = 5x Chain Rule. Suppose f and g have derivatives. Then (g f)'(x) = g'(f(x)) f '(x). We will see this in many guises. Theorem (Power Chain Rule) For any integer, Dx[ [f(x)] n ] = n [f(x)] n-1 Dx[f(x)] In words, to find the derivative of an expression to any power, find old power * expression to one lower power * derivative of the expression

5 5 Example. If k(x) = (3x 2 +1) 10, find k'(x) By the Power Chain Rule with n = 10, f(x) = (3x 2 +1), we find k '(x) = 10 (3x 2 +1) 9 Dx(3x 2 +1) = 10 (3x 2 +1) 9 (6x) = 60 x(3x 2 +1) 9 Example. Let g(x) = (x 3 + 2) 4 (a) Find g'(x). (b) Find the equation of the line tangent to the curve y = g(x) when x = 1. Solutions. (a) By the Power Chain Rule, g'(x) = 4 (x 2 +2) 3 (2x) = 8x (x 2 +2) 3 (b) When x = 1, y = g(1) = (1 3 +2) 4 = 3 4 = 81. the slope is g'(1) = 8(1)(1 2 +2) 3 = 8(3) 3 = 216. Hence the line is y = y1 + m(x-x1) y = (x-1) y = x y = 216 x Example. Find d/dt [ 3/(t 2 +1) 4 ] d/dt [ 3/(t 2 +1) 4 ] is the derivative of 3 (t 2 +1) -4 = d/dt [ 3(t 2 +1) -4 ] = 3 (-4) (t 2 +1) -5 Dt(t 2 +1) =-12 (t 2 +1) -5 (2t) =-24 t (t 2 +1) -5 We can combine the Power Chain Rule with other rules: Example. Find Dx [ x 2 (x 3 + 2) 5 ]. Since the expression is a product, we first use the product rule: Dx [ x 2 (x 3 + 2) 5 ]. = x 2 Dx [ (x 3 + 2) 5 ].+ (x 3 + 2) 5 Dx [ x 2 ]. = x 2 (5)(x 3 + 2) 4 Dx[x 3 +2].+ (x 3 + 2) 5 (2x) = 5x 2 (x 3 + 2) 4 (3x 2 ).+ (x 3 + 2) 5 (2x) = 15x 4 (x 3 + 2) 4.+ 2x(x 3 + 2) 5 Example. Find the derivative of x 2 (x 2 +2) 3 Solution This is (x 2 +2) 3 Dx (x 2 ) - x 2 Dx[(x 2 +2) 3 ] [(x 2 +2) 3 ] 2 = (x 2 +2) 3 (2x) - x 2 3(x 2 +2) 2 Dx[x 2 +2] [(x 2 +2) 3 ] 2

6 6 = (x 2 +2) 3 (2x) - x 2 3(x 2 +2) 2 (2x) (x 2 +2) 6 = (x 2 +2) 3 (2x) - 6x 3 (x 2 +2) 2 (x 2 +2) 6 = (x 2 +2) 2 [(x 2 +2) (2x) - 6x 3 ] (x 2 +2) 6 = [(x 2 +2) (2x) - 6x 3 ] (x 2 +2) 4 = [2x 3 +4x - 6x 3 ] (x 2 +2) 4 = -4x 3 +4x (x 2 +2) 4 Example. Find Dx [ (x 3 +2) 4 (x 2 +3) 5 ] We use the product rule: Dx [ (x 3 +2) 4 (x 2 +3) 5 ] = (x 3 +2) 4 Dx [ (x 2 +3) 5 ]+ (x 2 +3) 5 Dx [ (x 3 +2) 4 ] = (x 3 +2) 4 5 (x 2 +3) 4 Dx [ x 2 +3]+ (x 2 +3) 5 4(x 3 +2) 3 Dx [ x 3 +2] = (x 3 +2) 4 5 (x 2 +3) 4 (2x)+ (x 2 +3) 5 4(x 3 +2) 3 (3x 2 ) = 10x(x 3 +2) 4 (x 2 +3) x 2 (x 2 +3) 5 (x 3 +2) 3 The use of the Power Chain Rule also leads to the following formula: Theorem Dx [ x n ] = n x n-1 for all constants n. Example. Dx [x 5/7 ] = (5/7) x (5/7)-1 = (5/7) x -(2/7) Expressions written with surds are just powers written in a funny way. To find their derivatives, rewrite them in terms of their powers. Example. Dx [ x] = Dx [x 1/2 ] = (1/2) x -1/2 Example. Dx [ cube root of x 2 ] = Dx [ x 2/3 ]

7 7 = (2/3) x (2/3)-1 = (2/3) x (-1/3) The power chain rule also works in this setting with constant exponents Theorem (Power Chain Rule) For any constant n, Dx[ [f(x)] n ] = n [f(x)] n-1 Dx[f(x)] Example. Dx [ (x 3 +4)] = Dx [(x 3 +4) (1/2) ] = (1/2) (x 3 +4) (1/2)-1 Dx(x 3 +4) = (1/2) (x 3 +4) (-1/2) 3x 2 = (3/2)x 2 (x 3 +4) (-1/2) Example. Dx [ x(x 2 +1) 1/4 ] = x Dx [ (x 2 +1) 1/4 ] + (x 2 +1) 1/4 Dx[x] = x (1/4) (x 2 +1) (1/4)-1 Dx [ x 2 +1] + (x 2 +1) 1/4 = x (1/4) (x 2 +1) (-3/4) (2x) + (x 2 +1) 1/4 = (x 2 /2) (x 2 +1) (-3/4) + (x 2 +1) 1/4 Example. Dt [ 4 t (t 3 ) ] = Dt [ 4 t (t 3/2 ) ] = Dt [ 4 (t 5/2 ) ] = 4 (5/2) t (5/2)-1 = 10 t 3/2 You may think that the functions we have described are unlikely to occur in nature. Here are some examples. It turns out that quantities can be related in complicated ways. Example. Suppose that the population of bacteria at time t is P(t). It turns out in some models that R(t) = 0.3 P(t) - [P(t)] 2 / 500 is important since it is related to some additional nutriment that must be provided to the colony to permit healthy growth. Suppose that at time t = 5 we know P(5) = 120, P'(5) = 36. (a) Find R(5). (b) Find R'(5). (a) R(5) = 0.3 P(5) - [P(5)] 2 / 500 = 0.3 (120) - [120] 2 / 500 = 7.2 (b) R' (t) = Dt [ 0.3 P(t) - [P(t)] 2 / 500 ] = 0.3 Dt [ P(t)] - (1/500) Dt [[P(t)] 2 ] = 0.3 Dt [ P(t)] - (1/500) Dt [[P(t)] 2 ] = 0.3 P'(t) - (1/500)(2) P(t) Dt[P(t)] using the Power Chain Rule. = 0.3 P'(t) - (2/500) P(t) P'(t) Hence when t = 5 we know

8 8 R'(5) = 0.3 P'(5) - (2/500) P(5) P'(5) = 0.3(36) - (2/500) (120) (36) = Example. A model for the spruce budworm population u(t) involves the quantity G(t) = [u(t)] 2 1+ [u(t)] 2 [Murray, p. 33] Suppose u(1) = 10 and u'(1) = Find G(1) and G'(1). Solution: G(1) = 10 2 / ( ) = 100/101. Note that G'(t) means Dt[G(t)]. G'(t) = [1+[u(t)] 2 ] 2 u(t) u'(t) - [u(t)] 2 (2) u(t) u'(t) [1+[u(t)] 2 ] 2 Hence G'(1) = [1+[u(1)] 2 ] 2 u(1) u'(1) - [u(1)] 2 (2) u(1) u'(1) [1+[u(1)] 2 ] 2 G'(1) = [1+[10] 2 ] 2 (10) (0.61) - [10] 2 (2) (10) (0.61) [1+[10] 2 ] 2 G'(1) = (101) (20) (0.61) - (100) (20) (0.61) [101] 2 G'(1) = Example. A model for the concentration x(t) of carbon dioxide in the blood at time t involves the quantity V(t) = x(t) x(t) 0.3 [Murray p. 34] When t = 2, x = 0.6 and x' = Find V'(2). By the Quotient Rule, V'(t) = (1+x(t) 0.3 ) Dt [ x(t) 1.3 ] - x(t) 1.3 Dt [1+x(t) 0.3 ] (1+x(t) 0.3 ) 2 = (1+x(t) 0.3 ) (1.3) x(t) 0.3 Dt [ x(t)] - x(t) 1.3 (0.3) x(t) -0.7 Dt [x(t)] (1+x(t) 0.3 ) 2 = (1+x(t) 0.3 ) (1.3) x(t) 0.3 ( x'(t)) - x(t) 1.3 (0.3) x(t) -0.7 (x'(t)) (1+x(t) 0.3 ) 2 V'(2) = (1+(0.6) 0.3 ) (1.3) (0.6) 0.3 ( 0.02) - (0.6) 1.3 (0.3) (0.6) -0.7 (0.02) (1+(0.6) 0.3 ) 2 =

9 9 Arguments for why some of the quick rules are true: The "quick rules" above may initially look strange. So it is good to see why they are in fact true. Theorem (Product Rule) Dx [ f(x) g(x) ] = f(x) Dx[g(x)] + g(x) Dx[f(x)]. Idea of the argument: It depends on a clever way of adding 0. Dx [ f(x) g(x) ] = lim f(x+h)g(x+h) - f(x)g(x) h 0 h = lim f(x+h)g(x+h) -f(x+h)g(x)+f(x+h)g(x)- f(x)g(x) h 0 h = lim f(x+h)[g(x+h) -g(x)] + g(x)f(x+h)- g(x)f(x) h 0 h h = lim f(x+h)[g(x+h) -g(x)] + lim g(x)[f(x+h)- f(x)] h 0 h h 0 h = f(x) g'(x) + g(x) f '(x) Theorem. If f(x) = x k, where k is a negative integer then f ' (x) = k x k-1. Why is this true? f(x) = 1 x -k where -k>0 Hence by the quotient rule, f ' (x) = (x -k ) (0) - 1 (-k) x -k-1 (x -k ) 2 = k x -k-1 x -2k = k x (-k-1-(-2k)) = k x k-1 Theorem. (Chain Rule) (g f)'(x) = g'(f(x)) f '(x). Why is the Chain Rule true? Here is the rough idea: We wish to show that (g f)'(x) = g'(f(x)) f '(x). Since (g f) (x) = g(f(x)), we have (g f)'(x) = lim g(f(x+h)) - g(f(x)) h 0 h = lim g(f(x+h)) - g(f(x)) f(x+h) - f(x) h 0 h f(x+h)- f(x) [since we just multiplied by 1 in a curious form]

10 10 = lim g(f(x+h)) - g(f(x)) f(x+h) - f(x) h 0 f(x+h)- f(x) h We look at these two expressions, as h goes to 0, we know that f(x+h)-f(x) h goes to f'(x). For the first portion of the expression, we write k = f(x+h) - f(x) so that f(x+h) = f(x) + k. Now lim g(f(x+h)) - g(f(x)) h 0 f(x+h)- f(x) = lim g(f(x)+k) - g(f(x)) h 0 k = lim g(f(x)+k) - g(f(x)) k 0 k [since as h goes to 0, we expect k = f(x+h)-f(x) also to go to 0] = g'(f(x)) since g'(y) = lim g(y+h)-g(y) = lim g(y+k) - g(y) h 0 h k 0 k and this is the case where y = f(x). Hence we obtain that (g f)'(x) = g'(f(x)) f '(x) showing that the Chain Rule is true. Theorem. (Power Chain Rule) Dx[ [f(x)] n ] = n [f(x)] n-1 Dx[f(x)] We show that the Power Chain Rule arises from the general Chain Rule. Let g(x) = x n. Then Dx[ [f(x)] n ] = Dx[ g(f(x))] = g '(f(x)) f ' (x) But g'(x) = n x n-1, so g'(f(x)) = n [f(x)] n-1. Hence Dx[ [f(x)] n ] = n [f(x)] n-1 f ' (x) This proves the result, assuming the general chain rule. Theorem Dx [ x n ] = n x n-1 for all constants n. I give an argument in case n is a rational number, ie when n = p/q with p and q whole numbers (integers).

11 11 Then x n = x p/q So (x n ) q = (x p/q ) q = x p We use the chain rule with the equation (x n ) q = x p The Chain Rule says: (g f)'(x) = g'(f(x)) f '(x). We think f(x)= x n and g(x) = x q (because we recognize (x n ) q as the composition of functions.) We want to find f ' (x). But g'(x) = q x q-1 since q is an integer, so g '(f(x)) = q (f(x)) q-1 = q (x n ) q-1 = q (x (p/q) ) (q-1) = q x (p(q-1)/q) = q x (pq/q - p/q) = q x (p - p/q) g f(x) = (x n ) q = x p Since p is an integer, we have (g f)'(x) = p x p-1 Hence, from the Chain Rule, we obtain p x p-1 = q x (p(q-1)/q) f ' (x). This is an equation that can be solved for f'(x): f ' (x) = p x p-1 / [ q x (p - p/q) ] = (p/q) x p-1-p+p/q = (p/q) x p/q-1 = n x n-1

We first review various rules for easy differentiation of common functions: The same procedure works for a larger number of terms.

We first review various rules for easy differentiation of common functions: The same procedure works for a larger number of terms. 1 Math 182 Lecture Notes 1. Review of Differentiation To differentiate a function y = f(x) is to find its derivative f '(x). Another standard notation for the derivative is Dx(f(x)). Recall the meanings