3.5 Chain Rule 149 3.5 Chain Rule Introuction As iscusse in Section 3.2, the Power Rule is vali for all real number exponents n. In this section we see that a similar rule hols for the erivative of a power of a function y [g(x)] n. Before stating the formal result, let us consier an example when n is a positive integer. Suppose we wish to ifferentiate By writing (1) as y (x 5 1). (x 5 1), we can fin the erivative using the Prouct Rule: 2(x 5 1). 5x 4. Similarly, to ifferentiate the function y (x 5 1) 3, we can write it as y (x 5 1) 2. (x 5 1) an use the Prouct Rule an the result given in (2): 3(x 5 1) 2. 5x 4. x x n nx n1 y (x 5 1) 2. x (x5 1) 2 (x 5 1). x (x5 1) (x 5 1). x (x5 1) (x 5 1). 5x 4 (x 5 1). 5x 4 x (x5 1) 3 x (x5 1) 2. (x 5 1) we know this from (2) (x 5 1) 2. x (x5 1) (x 5 1). x (x5 1) 2 (x 5 1) 2. 5x 4 (x 5 1). 2(x 5 1). 5x 4 In like manner, by writing y (x 5 1) 4 as y (x 5 1) 3. (x 5 1) we can reaily show by the Prouct Rule an (3) that x (x5 1) 4 4(x 5 1) 3. 5x 4. Power Rule for Functions Inspection of (2), (3), an (4) reveals a pattern for ifferentiating a power of a function g. For example, in (4) we see bring own exponent as a multiple erivative of function insie parentheses 4(x 5 1) 3. 5x 4 c ecrease exponent by 1 For emphasis, if we enote a ifferentiable function by x [ ]n n[ ] n1 x [ ]. [ ], it appears that he foregoing iscussion suggests the result state in the next theorem. (1) (2) (3) (4) heorem 3.5.1 Power Rule for Functions If n is any real number an u g(x) is ifferentiable at x, then x [g(x)]n n[g(x)] n1. g (x), or equivalently, (6) x un nu n1. u x. (5)
150 CHAPER 3 he Derivative heorem 3.5.1 is itself a special case of a more general theorem, calle the Chain Rule, which will be presente after we consier some examples of this new power rule. EXAMPLE 1 Power Rule for Functions Differentiate y (4x 3 3x 1) 7. Solution With the ientification that u g(x) 4x 3 3x 1, we see from (6) that n u n1 u>x y x 7(4x3 3x 1) 6. x (4x3 3x 1) 7(4x 3 3x 1) 6 (12x 2 3). { EXAMPLE 2 Power Rule for Functions o ifferentiate y 1>(x 2 1), we coul, of course, use the Quotient Rule. However, by rewriting the function as y (x 2 1) 1, it is also possible to use the Power Rule for Functions with n 1: EXAMPLE 3 y x (1)(x2 1) 2. x (x2 1) (1)(x 2 1) 2 2x 2x (x 2 1) 2. Power Rule for Functions 1 Differentiate y (7x 5 x 4 2) 10. Solution Write the given function as y (7x 5 x 4 2) 10. Ientify u 7x 5 x 4 2, n 10 an use the Power Rule (6): y x 10(7x5 x 4 2) 11. x (7x5 x 4 2) 10(35x4 4x 3 ) (7x 5 x 4 2) 11. EXAMPLE 4 Power Rule for Functions Differentiate y tan 3 x. Solution For emphasis, we first rewrite the function as y (tan x) 3 an then use (6) with u tan x an n 3: y x 3(tan x)2. tan x. x Recall from (6) of Section 3.4 that (>x)tan x sec 2 x. Hence, y x 3 tan2 x sec 2 x. EXAMPLE 5 Quotient Rule then Power Rule Differentiate y (x2 1) 3 (5x 1) 8. Solution We start with the Quotient Rule followe by two applications of the Power Rule for Functions: Power Rule for Functions y (5x 1)8. x x (x2 1) 3 (x 2 1) 3. (5x 1)8 x (5x 1) 16 (5x 1)8. 3(x 2 1) 2. 2x (x 2 1) 3. 8(5x 1) 7. 5 (5x 1) 16
3.5 Chain Rule 151 6x(5x 1)8 (x 2 1) 2 40(5x 1) 7 (x 2 1) 3 (5x 1) 16 (x2 1) 2 (10x 2 6x 40) (5x 1) 9. EXAMPLE 6 Differentiate Solution Power Rule then Quotient Rule y A 2x 3 8x 1. By rewriting the function as 2x 3 y Q 8x 1 R 1>2 we can ientify an n 1 2. hus in orer to compute u>x in (6) we must use the Quotient Rule: y x 1 3 2 8x 1 R 1>2. 3 x 8x 1 R 1 3 2 8x 1 R 1>2. (8x 1). 2 (2x 3). 8 (8x 1) 2 1 3 2 8x 1 R 1>2. 26 (8x 1) 2. Finally, we simplify using the laws of exponents: u y x 13 (2x 3) 1>2 (8x 1) 3>2. 2x 3 8x 1 Chain Rule A power of a function can be written as a composite function. If we ientify f (x) x n an u g(x), then f (u) f (g(x)) [g(x)] n. he Chain Rule gives us a way of ifferentiating any composition f g of two ifferentiable functions f an g. heorem 3.5.2 Chain Rule If the function f is ifferentiable at u g(x), an the function g is ifferentiable at x, then the composition y ( f g)(x) f (g(x)) is ifferentiable at x an x f (g(x)) f (g(x)). g (x) y y or equivalently, (8) x u. u x. (7) PROOF FOR u 0 In this partial proof it is convenient to use the form of the efinition of the erivative given in (3) of Section 3.1. For x 0, u g(x x) g(x) or g(x x) g(x) u u u. In aition, y f (u u) f (u) f (g(x x)) f (g(x)). When x an x x are in some open interval for which u 0, we can write y y x u. u x. (9)
152 CHAPER 3 he Derivative Since g is assume to be ifferentiable, it is continuous. Consequently, as x S 0, g(x x) S g(x), an so from (9) we see that u S 0. hus, From the efinition of the erivative, (3) of Section 3.1, it follows that he assumption that u 0 on some interval oes not hol true for every ifferentiable function g. Although the result given in (7) remains vali when u 0, the preceing proof oes not. It might help in the unerstaning of the erivative of a composition y f (g(x)) to think of f as the outsie function an u g(x) as the insie function. he erivative of y f (g(x)) f (u) is then the prouct of the erivative of the outsie function (evaluate at the insie function u) an the erivative of the insie function (evaluate at x): erivative of outsie function x f (u) f (u). u. c erivative of insie function he result in (10) is written in various ways. Since y f (u), we have f (u) y>u, an of course u u>x. he prouct of the erivatives in (10) is the same as (8). On the other han, if we replace the symbols u an u in (10) by g(x) an g (x) we obtain (7). Proof of the Power Rule for Functions As note previously, a power of a function can be written as a composition of ( f g)(x) where the outsie function is y f (x) x n an the y n1 insie function is u g(x). he erivative of the insie function y f (u) u n is nu x u an the erivative of the outsie function is he prouct of these erivatives is then x. y y x u. u x nu n1 u x n[g(x)]n1 g (x). his is the Power Rule for Functions given in (5) an (6). rigonometric Functions We obtain the erivatives of the trigonometric functions compose with a ifferentiable function g as another irect consequence of the Chain Rule. For example, if y sin u, where u g(x), then the erivative of y with respect to the variable u is Hence, (8) gives or equivalently, y y lim Q lim x xs0 u R. u Q lim xs0 x R xs0 y Q lim us0 u R. u Q lim xs0 x R. y y x u. u x. y cos u. u y y x u. u x cos u u x. x sin[ ] cos[ ] x [ ] Similarly, if y tan u where u g(x), then y>u sec 2 u an so y y x u. u x u sec2 u x. note that u S 0 in the first term We summarize the Chain Rule results for the six trigonometric functions. (10)
3.5 Chain Rule 153 heorem 3.5.3 Derivatives of rigonometric Functions If u g(x) is a ifferentiable function, then x sin u cos u u x, x tan u u sec2 u x, x secu sec u tan u u x, x cos u sin u u x, x cot u u csc2 u x, x csc u csc u cot u u x. (11) (12) (13) EXAMPLE 7 Chain Rule Differentiate y cos 4x. Solution he function is cos u with u 4x. From the secon formula in (11) of heorem 3.5.3 the erivative is y u u x y x sin 4x. 4x 4 sin 4x. x EXAMPLE 8 Chain Rule Differentiate y tan(6x 2 1). Solution he function is tan u with u 6x 2 1. From the first formula in (12) of heorem 3.5.3 the erivative is u sec 2 u x y x sec2 (6x 2 1). x (6x2 1) 12x sec 2 (6x 2 1). EXAMPLE 9 Prouct, Power, an Chain Rule Differentiate y (9x 3 1) 2 sin 5x. Solution We first use the Prouct Rule: y x (9x3 1) 2. x sin 5x sin 5x. x (9x3 1) 2 followe by the Power Rule (6) an the first formula in (11) of heorem 3.5.3, from (11) from (6) y x (9x3 1) 2. cos 5x. x 5x sin 5x. 2(9x 3 1). (9x 3 1) 2. 5 cos 5x sin 5x. 2(9x 3 1). x (9x3 1) 27x 2 (9x 3 1)(45x 3 cos 5x 5 cos 5x 54x 2 sin 5x). In Sections 3.2 an 3.3 we saw that even though the Sum an Prouct Rules were state in terms of two functions f an g, they were applicable to any finite number of ifferentiable functions. So too, the Chain Rule is state for the composition of two functions f an g but we can apply it to the composition of three (or more) ifferentiable functions. In the case of three functions f, g, an h, (7) becomes x f (g(h(x))) f (g(h(x))). x g(h(x)) f (g(h(x))). g (h(x)). h (x).
154 CHAPER 3 he Derivative EXAMPLE 10 Repeate Use of the Chain Rule Differentiate y cos 4 (7x 3 6x 1). Solution For emphasis we first rewrite the given function as y [cos(7x 3 6x 1)] 4. Observe that this function is the composition ( f g h)(x) f (g(h(x))) where f (x) x 4, g(x) cosx, an h(x) 7x 3 6x 1. We first apply the Chain Rule in the form of the Power Rule (6) followe by the secon formula in (11): y x 4[cos (7x3 6x 1)] 3. x cos (7x3 6x 1) 4 cos 3 (7x 3 6x 1). csin(7x 3 6x 1). x (7x3 6x 1) 4(21x 2 6) cos 3 (7x 3 6x 1) sin (7x 3 6x 1). In the final example, the given function is a composition of four functions. first Chain Rule: ifferentiate the power secon Chain Rule: ifferentiate the cosine EXAMPLE 11 Differentiate Repeate Use of the Chain Rule y sin( tan23x 2 4). Solution he function is f (g(h(k(x)))), where f (x) sin x, g(x) tan x, h(x) 1x, an k(x) 3x 2 4. In this case we apply the Chain Rule three times in succession y x cos Atan23x 2 4 B. x tan23x2 4 cos Atan23x 2 4 B. sec 2 23x 2 4. x 23x2 4 cos Atan23x 2 4 B. sec 2 23x 2 4. x (3x2 4) 1>2 cos Atan23x 2 4 B. sec 2 23x 2 4. 1 2 (3x2 4) 1>2. x (3x2 4) first Chain Rule: ifferentiate the sine secon Chain Rule: ifferentiate the tangent rewrite power thir Chain Rule: ifferentiate the power cos Atan23x 2 4 B. sec 2 23x 2 4. 1 2 (3x2 4) 1>2. 6x 3x cos Atan23x2 4 B. sec 2 23x 2 4. 23x 2 4 simplify You shoul, of course, become so aept at applying the Chain Rule that you will not have to give a moment s thought as to the number of functions involve in the actual composition. x NOES FROM HE CLASSROOM (i) Probably the most common mistake is to forget to carry out the secon half of the Chain Rule, namely the erivative of the insie function. his is the u>x part in y y u x u x. 56 For instance, the erivative of y (1 x) 57 is not y>x 57(1 x) since 57(1 x) 56 is only the y>u part. It might help to consistently use the operation symbol >x: x (1 x) 57 57(1 x) 56. x (1 x) 57(1 x) 56. (1).
3.5 Chain Rule 155 (ii) A less common but probably a worse mistake than the first is to ifferentiate insie the given function. A stuent wrote on an examination paper that the erivative of y cos (x 2 1) was y>x sin (2x); that is, the erivative of the cosine is the negative of the sine an the erivative of x 2 1 is 2x. Both observations are correct, but how they are put together is incorrect. Bear in min that the erivative of the insie function is a multiple of the erivative of the outsie function. Again, it might help to use the operation symbol >x. he correct erivative of y cos (x 2 1) is the prouct of two erivatives. y x sin (x 2 1). x (x 2 1) 2x sin (x 2 1).