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Lesson-5 Introduction to Calculus Chain Rule The chain rule applies to procedures when one of the functions is a composition of other function. A function of a function can be regarded as a function within a function, such as, for example, y = (x2 + 1)8 It is inconvenient, and sometimes impossible, to expand such functions into polynomials. The way to tackle the problem is to substitute u for the inner function; that is, substitute u = x2 + 1 (5.1) so that y = u8 (5.2) As x increases to x + x there will be corresponding increases of u and y in u and y respectively. or means multiplication. Since these finite quantities it is possible to write down: dy dy du dx dx du or better in different order as y y u u 1 u x (because u x ) Both y and u are dependent variables as x 0 and u 0 , then y dy x dx y dy u du u du x dx Therefore, in the limit this becomes the chain rule dy dy du dx du dx (5.3) Returning to the function y = (x2 + 1)8 and substitution u = x2 + 1 du = 2x dx and dy = 8x7 du so that dy dy du = 8u7*2x = 8(x2 + 1)2x = 16x(x2 + 1)7 dx du dx Example 1 Differentiate y = (x2 - 6x + 1)1/2 Solution: let u = x2 - 6x + 1 so that du = 2x - 6 dx then y =u1/2 dy = 1/2u-1/2 du and dy dy du = 1/2u-1/2(2x – 6) = dx du dx ½(x2 – 6x + 1)-1/2(2x – 6) if we taking 2 in front of second bracket, we obtain: (x – 3)(x2 – 6x + 1)-1/2 (because ½ and 2 cancel each other) Differentiation of a function of a function may be carried out without going through this substitution as procedure each time. The method can be summarised as shown below: a) Differentiate the first function as a simple variable. b) Differentiate the internal function with respect to x. c) dy = product (multiplication) of (a) and (b). dx Very often the chain rule involves more than two functions. In this case procedure is similar to describe above, but more function are involved. EXERCISES: Differentiate the following with respect to appropriate variable. 1. (2x + 3)4 2. (7x2 – 3)6 3. (-3x1/2 + 4)3 4. (4t – 3)1/2 5. (6t2 - 62 )3/4 6. (Q - 1 )2 t 1 7. 3 x 3 Q 1 8. t 3t 4 2 2 9. (3x + 2)6 10. (5 – x2)3 11. 3(x5 – 8x2 + x)100 12. (x2 – 2)-3 13. (2x2 – 3x + 1)-10/3 14. 5x 2 x 15. 4 2x 1 16. 2 1 ( x 3x) 2 17. 6 2x x 1 18. 19. 2( 8x 1 )-1 20. 2 ( x 3 1) 2 5 2 3 7x + 3 7x Answers: 1. 8(2x + 3)3 2. 4. (4t - 3)-1/2 5. 84x(7x2 – 3)5 3. -9/2x-1/2(3x1/2 + 4)2 9 (t + 1 / t3) (6t2 - 6/t2)-1/4 6. (Q - 1/Q)(1 + 1/Q2) 7. -2/3x(x2 – 3)-4/3 8. -1/2(2t – 3)(t2 – 3t + 4)-3/2 9. 18 (3x + 2)5 10. -6x(5 – x2)2 11. 12. -6x(x2 – 2)-4 13. 300 (x5 – 8x2 +x)99(5x4 - 16x +1) -10/3(4x - 3)(2x2 3x + 1)-13/3 14. 1/2(10x - 1)(x2 –x)-1/2 15. ½(2x – 1)-3/4 -6(x – 1)(2x2 – x + 1)-2 16. 12/5x2(x3 + 1)-3/5 17. 18. -2(2x - 3)(x2 - 3x)-3 20. 2/3(7)1/3 x-2/3 = 14/9x-2/3 19. -8(8x - 1)-3/2