SECT ION 3.2 The Derivative as a Function 201 SOLUTION Figure (A) satisfies the inequality f .a h/ f .a h/ f .a h/ f .a/ C C 2h h since in this graph the symmetric difference quotient has a larger negative slope than the ordinary right difference quotient. [In figure (B), the symmetric difference quotient has a larger positive slope than the ordinary right difference quotient and therefore does not satisfy the stated inequality.] 75. Show that if f .x/ is a quadratic polynomial, then the SDQ at x a (for any h 0) is equal to f 0.a/ . Explain the graphical meaning of this result. D ¤ SOLUTION Let f .x/ px 2 qx r be a quadratic polynomial. We compute the SDQ at x a. D C C D f .a h/ f .a h/ p.a h/ 2 q.a h/ r .p.a h/ 2 q.a h/ r/ C C C C C C C 2h D 2h pa2 2pah ph 2 qa qh r pa 2 2pah ph 2 qa qh r C C C C C C C D 2h 4pah 2qh 2h.2pa q/ C C 2pa q D 2h D 2h D C Since this doesn’t depend on h, the limit, which is equal to f 0.a/ , is also 2pa q. Graphically, this result tells us that the secant line to a parabola passing through points chosen symmetrically about x a is alwaysC parallel to the tangent line at x a. D D 76. Let f .x/ x 2. Compute f .1/ by taking the limit of the SDQs (with a 1) as h 0. D 0 D ! SOLUTION Let f .x/ x 2. With a 1, the symmetric difference quotient is D D 1 1 2 2 f .1 h/ f .1 h/ .1 h/2 .1 h/2 .1 h/ .1 h/ 4h 2 C C C : 2h D 2h D 2h.1 h/2.1 h/ 2 D 2h.1 h/2.1 h/ 2 D .1 h/2.1 h/ 2 C C C Therefore, 2 f 0.1/ lim 2: D h 0 .1 h/2.1 h/ 2 D ! C 3.2 The Derivative as a Function Preliminary Questions 1. What is the slope of the tangent line through the point .2; f .2// if f .x/ x3? 0 D SOLUTION The slope of the tangent line through the point .2; f .2// is given by f .2/ . Since f .x/ x3, it follows that 0 0 D f .2/ 23 8. 0 D D 2. Evaluate .f g/ 0.1/ and .3f 2g/ 0.1/ assuming that f 0.1/ 3 and g0.1/ 5. C D D SOLUTION .f g/ .1/ f .1/ g .1/ 3 5 2 and .3f 2g/ .1/ 3f .1/ 2g .1/ 3.3/ 2.5/ 19 . 0 D 0 0 D D C 0 D 0 C 0 D C D 3. To which of the following does the Power Rule apply? (a) f .x/ x2 (b) f .x/ 2e D D (c) f .x/ xe (d) f .x/ ex D D (e) f .x/ xx (f) f .x/ x 4=5 D D SOLUTION (a) Yes. x2 is a power function, so the Power Rule can be applied. (b) Yes. 2e is a constant function, so the Power Rule can be applied. (c) Yes. xe is a power function, so the Power Rule can be applied. (d) No. ex is an exponential function (the base is constant while the exponent is a variable), so the Power Rule does not apply. (e) No. xx is not a power function because both the base and the exponent are variable, so the Power Rule does not apply. 4=5 (f) Yes. x is a power function, so the Power Rule can be applied. 4. Choose (a) or (b). The derivative does not exist if the tangent line is: (a) horizontal (b) vertical. SOLUTION The derivative does not exist when: (b) the tangent line is vertical. At a horizontal tangent, the derivative is zero. 202 CHAPTER 3 DIFFERENTIATION 5. Which property distinguishes f .x/ ex from all other exponential functions g.x/ bx? D D SOLUTION The line tangent to f .x/ ex at x 0 has slope equal to 1. D D Exercises In Exercises 1–6, compute f 0.x/ using the limit definition. 1. f .x/ 3x 7 D SOLUTION Let f .x/ 3x 7. Then, D f .x h/ f .x/ 3.x h/ 7 .3x 7/ 3h f 0.x/ lim C lim C lim 3: D h 0 h D h 0 h D h 0 h D ! ! ! 2. f .x/ x2 3x D C SOLUTION Let f .x/ x2 3x . Then, D C f .x h/ f .x/ .x h/ 2 3.x h/ .x 2 3x/ f 0.x/ lim C lim C C C C D h 0 h D h 0 h ! ! 2xh h2 3h lim C C lim .2x h 3/ 2x 3: D h 0 h D h 0 C C D C ! ! 3. f .x/ x3 D SOLUTION Let f .x/ x3. Then, D f .x h/ f .x/ .x h/ 3 x3 x3 3x 2h 3xh 2 h3 x3 f 0.x/ lim C lim C lim C C C D h 0 h D h 0 h D h 0 h ! ! ! 3x 2h 3xh 2 h3 lim C C lim .3x 2 3xh h2/ 3x 2: D h 0 h D h 0 C C D ! ! 4. f .x/ 1 x 1 D SOLUTION Let f .x/ 1 x 1. Then, D 1 1 .x h/ x 1 x h 1 x C f .x h/ f .x/ x.x h/ 1 1 f 0.x/ lim C lim C lim C lim : D h 0 h D h 0 h D h 0 h D h 0 x.x h/ D x2 ! ! ! ! C 5. f .x/ x px D SOLUTION Let f .x/ x px. Then, D f .x h/ f .x/ x h px h .x px/ px h px px h px f 0.x/ lim C lim C C 1 lim C C C D h 0 h D h 0 h D h 0 h px h px ! ! ! ! C C .x h/ x 1 1 1 lim C 1 lim 1 : D h 0 h. px h px/ D h 0 px h px D 2px ! C C ! C C 6. f .x/ x 1=2 D SOLUTION Let f .x/ x 1=2 . Then, D 1 1 p f .x h/ f .x/ px h px px x h f 0.x/ lim C lim C lim C D h 0 h D h 0 h D h 0 hpx hpx ! ! ! C Multiplying the numerator and denominator of the expression by px px h, we obtain: C C px px h px px h x .x h/ f 0.x/ lim C C C lim C D h 0 hpx hpx px px h D h 0 hpx hpx. px px h/ ! C C C ! C C C 1 1 1 lim : D h 0 px hpx. px px h/ D pxpx.2 px/ D 2x px ! C C C SECT ION 3.2 The Derivative as a Function 203 In Exercises 7–14, use the Power Rule to compute the derivative. d 7. x4 dx ˇx 2 ˇ D ˇ d 4 3 d 4 3 SOLUTION ˇ x 4x so x 4. 2/ 32 . dx D dx D D ˇx 2 ˇ D d ˇ 8. t 3 ˇ dt ˇt 4 ˇ D ˇ d 3 4 d 3 4 3 SOLUTION ˇ t 3t so t 3.4/ . dt D dt D D 256 ˇt 4 ˇ D d ˇ 9. t2=3 ˇ dt ˇt 8 ˇ D ˇ d 2=3 2 1=3 d 2=3 2 1=3 1 SOLUTION ˇ t t so t .8/ . dt D 3 dt D 3 D 3 ˇt 8 ˇ D d ˇ 10. t 2=5 ˇ dt ˇt 1 ˇ D ˇd 2=5 2 7=5 d 2=5 2 7=5 2 SOLUTION ˇ t t so t .1/ . dt D 5 dt D 5 D 5 ˇt 1 ˇ D d ˇ 11. x0:35 ˇ dx d 0:35 0:35 1 0:65 SOLUTION x 0:35.x / 0:35x : dx D D d 12. x14=3 dx d 14=3 14 .14=3/ 1 14 11=3 SOLUTION x x x : dx D 3 D 3 d 13. tp17 dt d p17 p17 1 SOLUTION t p17t dt D d 2 14. t dt d 2 2 2 1 SOLUTION .t / t dt D In Exercises 15–18, compute f .x/ and find an equation of the tangent line to the graph at x a. 0 D 15. f .x/ x4, a 2 D D 4 3 SOLUTION Let f .x/ x . Then, by the Power Rule, f 0.x/ 4x . The equation of the tangent line to the graph of f .x/ at x 2 is D D D y f .2/.x 2/ f .2/ 32.x 2/ 16 32x 48: D 0 C D C D 16. f .x/ x 2, a 5 D D 2 3 SOLUTION Let f .x/ x . Using the Power Rule, f 0.x/ 2x . The equation of the tangent line to the graph of f .x/ at x 5 is D D D 2 1 2 3 y f .5/.x 5/ f .5/ .x 5/ x : D 0 C D 125 C 25 D 125 C 25 17.