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.
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1 The chain rule 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 Just as in this example, usually g f and f g are not the same. Example. Write (2x+3) 4 as g f, by telling what is g and what is f. Solution. 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. Solution. g(x) = x since the last thing done is to take the square root. And f(x) = 5x The main result for differentiation of compositions of functions is called the "Chain Rule." Chain Rule. Suppose f and g have derivatives. Then (g f)'(x) = g'(f(x)) f '(x). We will see this in many guises. Following is the result where g(u) = u n, so (g f)(x) = (f(x)) n. Example. If k(x) = (3x 2 +1) 10, find k'(x) Solution. We recognize that k(x) = (g f)(x) where g(x)= x 10 and f(x) = (3x 2 + 1). Hence g '(x) = 10 x 9 and f '(x) = 6 x. By the Chain Rule k '(x) = (g f)'(x) = g'(f(x)) f '(x) = 10 (3x 2 +1) 9 (6x) = 60 x(3x 2 +1) 9. This kind of example occurs so often that there is an explicit formula for differentiating a power: Corollary. (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
2 31 We will now redo the prvious example, using the Power Chain Rule rather than the Chain Rule. The Power Chain Rule is a bit shorter and easier, but also one more thing to remember. Example. If k(x) = (3x 2 +1) 10, find k'(x) Solution. 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 (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 (3 x 2 ) = 12 x 2 (x 2 +2) 3 (b) When x = 1, y = g(1) = (1 3 +2) 4 = 3 4 = 81. the slope is g'(1) = 12(1)(1 2 +2) 3 = 12(3) 3 = 324. Hence the line is y = y1 + m(x-x1) y = (x-1) y = x y = 324 x Example. Find d/dt [ 3/(t 2 +1) 4 ] Solution. 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 Example. Let f(x) = (2 x 3 + 3) 7 (a) Find the derivative of f. (b) Find the instantaneous rate of change when x = 1 Solution. (a) f ' (x) = 7 (2 x 3 + 3) 7-1 Dx(2 x 3 + 3) = 7 (2 x 3 + 3) 6 (6 x 2 ) = 42 x 2 (2 x 3 + 3) 6 (b) The instantaneous rate of change is f '(1) = 656,250. Example. If f(t) = 7 (3 t 4 - t) 5, find f '(t). Solution. f '(t) = 7 (5) (3 t 4 - t) 4 Dt(3 t 4 - t) = 35 (3 t 4 - t) 4 (12 t 3-1) We can combine the Power Chain Rule with other rules: Example. Find Dx [ x 2 (x 3 + 2) 5 ]. Solution. Since the expression is a product, we first use the product rule:
3 32 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 + 1) 4 (2x 3 ) Solution. (x 2 + 1) 4 Dx(2x 3 ) + (2x 3 ) Dx[(x 2 + 1) 4 ] = (x 2 + 1) 4 (6x 2 ) + (2x 3 ) 4 (x 2 + 1) 3 Dx[(x 2 + 1)] = (x 2 + 1) 4 (6x 2 ) + (2x 3 ) 4 (x 2 + 1) 3 (2x) = (x 2 + 1) 4 (6x 2 ) + (16x 4 ) (x 2 + 1) 3 Example. Find the derivative of x (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 = (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) (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
4 33 = -4x 3 +4x (x 2 +2) 4 Example. Find Dx [ (x 3 +2) 4 (x 2 +3) 5 ] Solution. 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 It is not always easy to decide what is the simplest form for the answer, since what is simplest depends on the use that is to be made of the formula. In this case, we can factor out x from both terms, as well as (x 3 +2) 3 since both terms contain that many powers of (x 3 + 2), and (x 2 + 3) 4 since both terms contain that many powers of (x 2 + 3). We obtain 10x(x 3 +2) 4 (x 2 +3) x 2 (x 2 +3) 5 (x 3 +2) 3 = x (x 3 +2) 3 (x 2 + 3) 4 [ 10 (x 3 + 2) + 12 x (x 2 +3) ] = x (x 3 +2) 3 (x 2 + 3) 4 [ 10 x x x ] = x (x 3 +2) 3 (x 2 + 3) 4 [ 22 x x + 20 ] which is the best form for the answer. 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 ] = (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 exponent) For any constant n,
5 34 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 Problems for If f(x) = (2x 2 + 3) 8, find f '(x). 32 x (2x 2 +3) 7 2. If g(x) = (3x 3 + 2x + 1) 6, find g'(x). 6 (3x 3 + 2x + 1) 5 (9 x 2 + 2) 3. Find Dx [ (x 3 + 1) 5 ] 15 x 2 (x 3 + 1) 4 4. Suppose P(t) = (t 4 +1) 4 (a) Find P'(t). (b) Find the equation of the line tangent to y = P(t) at t = 1. (c) Find the instantaneous rate of change when t = 2. (a) 16 t 3 (t 4 +1) 3 (b) y = 128 t (c) 628, Suppose f (x) = x 3 (x 2 + 1) 4 (a) Find f '(x). (b) Find f '(1). (c) Find the equation of the line tangent to y = f(x) at x = 1. (a) 8 x 4 (x 2 + 1) x 3 (x 2 + 1) 4 (b) 112
6 35 (c) y = 112 x Suppose g(x) = 2x 2 (3x 3 + 1) 3 (b) Find g'(1). (a) 54 x 4 (3 x 3 + 1) x (3 x 3 + 1) 3 (b) Find the derivative: (a) (x 3-4x 2 + 5) 5 (b) 1/(2 t 2-3t) 4 (c) 2x 2 (3x+7) 4 (d) (x 4 + 3) (e) 5x/(2x+1) 3 (f) (x 2 + 3) 3/4 (a) 5 (x 3-4 x 2 + 5) 4 (3 x 2-8 x) (b) -4 (2t 2-3 t) -5 (4t -3) (c) 24 x 2 (3x+7) 3 + 4x (3x+7) 4 (d) 2 x 3 (x 4 + 3) -1/2 (e) (5-20 x) / (2x+1) 4 (f) (3x/2) (x 2 + 3) -1/4 8. Find the equation of the line tangent to the graph of y = 3/(x-2) 2 at x = 1. y = 6x-3 9. Suppose g(x) = 2x (3x+1) 3 (b) Find g'(1). (a) 12 x (3x+1) 4 (b) - 19/ Suppose f(x) = 2x (x-2) 2 (a) Find f '(x). (b) Find f '(1). (a) 2 x (x-2) 3 (b) 8
7 Suppose g(x) = 2 x (x 2 + 1) 4 (b) Find the slope of the line tangent to y = g(x) at x = 2. (a) x 2 (6-10 x 2 ) (x 2 + 1) 5 (b) / Suppose g(t) = 2 t t + 3 t (a) Find g'(t) (b) Find g'(4). (a) 4 t (3/2) t -1/2 (b) Suppose f(x) = (2 x 3 + 3) -4. (a) Find f ' (x). (b) Find the slope of the line tangent to y = f(x) where x = 1. (a) -24 x 2 (2 x 3 +3) -5 (b) Suppose g(x) = (3 x 2 + 1) -(2/3). (a) Find g ' (x). (b) Find the slope of the line tangent to y = g(x) where x = 2. (a) -4 x (3x 2 + 1) -(5/3) (b) Suppose g(x) = (4 x - 3) -(1/2). (a) Find g ' (x). (b) Find the slope of the line tangent to y = g(x) where x = 1. (a) -2 (4x - 3) -(3/2) (b) Suppose f(x) = (x 3 +1) Find f '(x). (3/2) x 2 (x 3 + 1) -1/2 = 3 x (x 3 + 1) 17. Suppose f(x) = (x 3 +1) 4/3 Find f '(x). 4 x 2 (x 3 + 1) 1/3 18. Suppose f(x) = x (2x 2 +1) Find f '(x). 2 x 2 (2 x 2 + 1) -(1/2) + (2 x 2 + 1) (1/2)
8 37 = 4 x (2 x 2 + 1) 19. Suppose g(x) = 2 x 2/3 + 3 x 4/3 (a) Find g'(x) (b) Find g'(8). (a) (4/3) x -(1/3) + 4 x (1/3) (b) 26/3 20. Suppose g(x) = 3 x (1/3) + 6 x (2/3) (b) Find g'(8). (c) Find the equation of the line tangent to y = g(x) at x = 8. (a) x -(2/3) + 4 x -(1/3) (b) 9/4 = 2.25 (c) y = 2.25 x Suppose g(x) = x 4 (x 2 + 3x + 1) 6 (b) Find g'(1). (a) 6 x 4 (x 2 + 3x + 1) 5 (2x + 3) + 4 x 3 (x 2 + 3x + 1) 6 (b) 156, Suppose g(x) = (x 2 + 1) 3 (x 2 + 2x + 3) 4 Find g'(x). 8 (x+1) (x 2 + 1) 3 (x 2 + 2x + 3) x (x 2 + 1) 2 (x 2 2x + 3) Suppose g(x) = (2x+1) (x 2 + 2x + 3) 3 Find g'(x). (2x+1) (6-10x - 8 x 2 ) (x 2 + 2x + 3) Suppose g(x) = (2x+1) (3x + 2) (b) Find g'(2) (a) (3x + 2) 3/2 (2x+1) 1/2 (b) Suppose P(t) = t 2 t (3/5). (a) Find P'(t). (b) Find the instantaneous rate of change of P when t = 32.
9 38 (a) (13/5) t (8/5) (b) 3328/5 = Suppose g(x) = (3x + 2) (x 2 + 4x + 1) 2 Find g'(x). 9 x x (x x + 1) Suppose P(t) = 6 t t (2/3). (a) Find P'(t). (b) Find the instantaneous rate of change of P when t = 8. (a) 8 t (1/3) (b) 16
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