Section 2.3 Rules for Finding Derivatives • EXAMPLE S Find DrV + 113 . SOLUTION Dry = ) 2 4 D( + Dr() 4 + 1)Dr(2) (x — 4 + 1) 2D(x 4 + 1) (x - — (x + i)(O) (2)(4x) + 0) (3)(1) 4 + 1)2 (x 2 x - —Sx 3 2 + 4 (x 1) 2 x — EXAMPLE 6 (x) 1 xDr(3) —31) + I Show that the Power Rule holds for negative integral expo nents; that is, ic” = D(x) Dr(i) _ We saw as part of Example 5 that D(3/x) way to see the same thing. n-I = x”O—1,ix’ = = “1 —3/x. Now we have another ( loncepts Review I. The derivative of a product of two functions is the first plus the lies times the derivative of the first. In q ihuls. D [f(x)g(x)] = 3. The second term (the term involving Ii) in the expansion of (x + h)’ is It is this fact that leads to the formula D, . 2. The derivative of a quotient is the times the derivaP’, of the numerator minus the numerator times the derivative 1 he all divided by the In symbols, 1’, f(v)/g(x)j = [i”] = 4. L is called a linear operator if L(kf) derivative operator denoted by LU + g) = such an operator. [ Problem Set 2.3 Iii Ii ,h1cnis 1—44, tind D,v its j,,t I. y = 3. 3’ S. y = 7. y = the nih’s of this section. 2 2x 2. lix 4. y 2 2x y = = TX 6. y = 8. y = 17. = 19. y = 2 no 10. y a x 2 + 13. v 14. v IS. y 2x 12. v x + x + 2 + = 3x 4 = irx 7 l — — 2r7 2x 16. yx’+5x 7 — 5 —2 — = 5x + irx + 5x —7 —ITX - —I)) -, 2x = 23. v = 2 + 1) x(x = (2x + 1)2 4x 25. 4 + x 3x 27. v= y — 2x + v (v + 30. y = 4 (x 31. y = 2 (5x 32. y = 4 (3x + 2x)(x + 1) — — 3 2x)(x 18. v = 20. y = 22. v = 24. v = 26. v t + 2 (x + 2)(x = 29. + x 21. v — 3a = 1 — I I 9. + 3 3x + 2 7)(3x 28. v 3x + 1) — — 2 2x + 1) 2x + 1) 3x + 1) 2x6 1 3 = x. 2 3x + — — 3x(x (—3x x - 2 3 — 1) + 2)2 + 1) 2 — I)(x 4 (x and is