1. 13. Functions and Derivatives Objectives: 1.Derivatives of (u(x)) n. 2.Derivation of the chain rule. 3.Examples. Refs: B&Z 10.3.

2. So far we have revised our differentiation techniques for simple functions (logs, exp, sin, polynomials) and used those techniques in slightly more complicated examples (sums, products, quotients). We will now look at slightly harder examples. The Chain Rule In some of our simple examples we have already used the chain rule but we probably didn’t realize it. As an introductory example we let u(x) be some function that we can differentiate, and we will differentiate (u(x)) n, where n is an integer.

3. First let n=2. We can differentiate (u(x)) 2 using the product rule. (u(x)) 2 = u(x).u(x) So ((u(x)) 2 )’= u’(x).u(x) + u(x).u’(x) = 2.u(x).u’(x). Now let n=3. (u(x)) 3 = (u(x)) 2. u(x) Again we can use the product rule. So ((u(x)) 3 )’= ((u(x)) 2 )’.u(x) + (u(x)) 2. u’(x)

4. ((u(x)) 3 )’= ((u(x)) 2 )’. u(x) + (u(x)) 2. u’(x) We already know that ((u(x)) 2 )’ = 2.u(x).u’(x), so ((u(x)) 3 )’= 2.u(x).u’(x).u(x) + (u(x)) 2. u’(x) = 3(u(x)) 2. u’(x). Now let n=4. (u(x)) 4 = (u(x)) 3. u(x) = ((u(x)) 3 )’. u(x) + (u(x)) 3. u’(x) ( product rule ) = 3(u(x)) 2. u’(x).u(x) + (u(x)) 3. u’(x) = 4(u(x)) 3. u’(x). We continue on in this fashion…….but the general rule is: ((u(x)) n )’= n (u(x)) n-1. u’(x)

5. Example 1: y=(5x 2 +3) 8 Here u(x) = 5x 2 +3 and so u’(x) = 10x. So y = (u(x)) 8 …….and applying the rule gives: dx dy = 8(u(x)) 7. u’(x). We must write the answer in terms of x not u(x). dx dy = 8(5x 2 +3 ) 7. 10x. = 80x(5x 2 +3 ) 7.

6. Example 2: y = sin 3 (x) Here u(x) = sin(x) and so u’(x) = cos(x). So y = (u(x)) 3 …….and applying the rule gives: dx dy = 3(u(x)) 2. u’(x). We must write the answer in terms of x not u(x). dx dy = 3(sin(x)) 2. cos(x).

7. To differentiate a composite function we must be able to break the function down into simpler pieces - an inner function and an outer function. In the previous type of example this is easy. x 5x 2 +3 (5x 2 +3) 8 inner function outer function u(x) (u(x)) 8 x sin(x) (sin(x)) 3 inner function outer function u(x) (u(x)) 3

8. Other examples: y = log e (x 3 + 3x 2 + 2) x x 3 + 3x 2 + 2 log e (x 3 + 3x 2 + 2) inner function outer function u(x) log e (u(x)) x 6x + 3 sin(6x + 3) inner function outer function u(x) sin(u(x)) y = sin(6x + 3)

9. In each of these examples we have written our composite function as m(x) = f(g(x)) where f and g are both functions that we can differentiate. Once we have written the function in this form we can use The Chain Rule to differentiate. Sketch Derivation Suppose that for some function y = m(x) its’ decomposition is m(x) = f(g(x)). Then

10. Now let g(x) = u and g(x+h) - g(x) = k. Then u+k = g(x) + g(x+h) - g(x) = g(x+h) Note that since k = g(x+h) - g(x), k approaches zero whenever h approaches zero. Now

11. In Leibniz notation this is Examples 1.f(x) = sin(3x) 2.f (x) = √(x 4 + 3x) 3.f(x) = log e (x 3 +2x+3) Solutions Remember that the first step in solving any of the problems is to identify the inner and outer functions.

12. 1. y=sin(3x) x 3x sin(3x) inner function outer function So we let u=3x. So y=sin(u), u=3x. By the chain rule, dy du. dx = cos(u). 3. =cos(u), du dx =3. dy du dy dx Therefore = 3cos(3x)

13. 2. y=√(x 4 + 3x) x x 4 + 3x √(x 4 + 3x) inner function outer function So we let u= x 4 + 3x.So y=√u, u= x 4 + 3x. By the chain rule, dy du. dx = 1/2 u -1/2.(4x 3 +3) =1/2 u -1/2, du dx = 4x 3 +3. dy du dy dx Therefore = 1/2 (x 4 + 3x ) -1/2.(4x 3 +3)

14. 3. y=log e (x 3 + 2x+3) x x 3 + 2x+3 log e (x 3 + 2x+3) inner function outer function So we let u= x 3 + 2x+3.So y=log e (u), u= x 3 + 2x+3. By the chain rule, dy du. dx = 1/u.(3x 2 +2) =1/u, du dx = 3x 2 +2. dy du dy dx Therefore = 3x 2 + 2 x 3 +2x+3

15. Some composite functions have more than two components - the principle for using the chain rule is still the same. Extended Chain Rule Suppose that m(x)=f 1 (f 2 (f 3 ……….(f n (x)))). Then m’(x)= f 1 ’(f 2 (f 3 ……….(f n (x)))). f 2 ’(f 3 ……….(f n (x)))…….. ……… f n-1 ’(f (x)).f’(x). Alternative notation: dy dx = du n-1 du n-2 dy du n-1 du 1 dx du 2 du 1 …………

16. Examples 1. y = sin(log e (x 3 +2)) So y=sin(u 2 ) and dy du 2 = cos (u 2 ) u 2 = log e (u 1 ) and du 2 du 1 =1/u 1 u 1 =x 3 +2du 1 dx = 3x 2 x x 3 +2 log e (x 3 +2) sin(log e (x 3 +2)) u1u1 u2u2

17. Now apply the chain rule: du 1 dx du 2 du 1 dy dx dy du 2 = = cos(u 2 ). 1/u 1. 3x 2 = cos(log e (u 1 )).1/(x 3 +2). 3x 2 = 3x 2. cos(log e (x 3 +2)) ( x 3 +2)

18. Examples 1. y=log e (2+exp(sin(x 3 +2x+3))) So y = log e (u 3 ) and dy du 3 = 1/u 3 u 2 = sin(u 1 ) and du 2 du 1 =cos(u 1 ) du 1 dx u 1 =x 3 +2x+3=3x 2 +2 x x 3 +2x+3 sin(x 3 +2x+3) 2+exp(sin(x 3 +2x+3)) log e (2+exp(sin(x 3 +2x+3)) u 3 u1u1 u2u2 u 3 = 2+exp(u 2 ) du 3 du 2 = exp(u 2 )

19. Now apply the chain rule: du 1 dx du 3 du 2 dy dx dy du 3 = du 2 du 1 = exp(sin(x 3 +2x+3)).cos(x 3 +2x+3).(3x 2 +2) 2+exp(sin(x 3 +2x+3)) =1/u 3. exp(u 2 ).cos(u 1 ).(3x 2 +2) = (1/(2+exp(u 2 ))). exp(sin(u 1 ) ).cos(x 3 +2x+3).(3x 2 +2) = (1/(2+exp(sin(u 1 )))). exp(sin(x 3 +2x+3)).cos(x 3 +2x+3).(3x 2 +2) =(1/(2+exp(sin(x 3 +2x+3)))). exp(sin(x 3 +2x+3)).cos(x 3 +2x+3).(3x 2 +2)

20. You should now be able to attempt Q’s 5,7 from Example Sheet 5 from the Orange Book.