M408C: Derivatives and Differentiation RulesSeptember 18, 2008Recall the definition of continuity: a function f is continuous at a number a if limx→af(x) =f(a). This definition implicitly requires three things to be true for f to be continuous at a:1.2.3.14. (3.1.52) Determine whether f0(0) exists forf(x) =x2sin1xx 6= 00 x = 0Solution: Using the definition of the derivative, we havef0(0) = limh→0f(0 + h) − f(0)h= limh→0h2sin1h− 0h= limh→0h sin1h.So we want to know if this limit exists. We will use the Squeeze Theorem by noting that−1 ≤ sin1h≤ 1, so −|h| ≤ |h| sin1h≤ |h|. We must use |h| since h could be negative or posi-tive. To deal with these absolute values, we will handle the cases of h > 0 and h < 0 separately.Assume that h > 0. Then |h| = h, so our inequality becomes simply −h ≤ h sin1h≤ h. Applyingthe Squeeze Theorem and noting that limh→0(−h) = limh→0h = 0, we see that limh→0h sin1h= 0.Now assume that h < 0. Then |h| = −h, so our inequality becomes h ≤ −h sin1h≤ −h.Multiplying through by −1, we have that −h ≥ h sin1h≥ h. Rewriting this as h ≤ h sin1h≤ −hand noting that h < 0 (so −h > 0), we see that we can apply the Squeeze Theorem again andconclude that limh→0h sin1h= 0.Thus the limit exists for h > 0 and h < 0, so the limit exists.5. (3.2.29) If f(x) = x4+ 2x, find f0(x). Check your answer by drawing the graphs of f and f0.I will leave the graphs to you once you have f0. We can use what we know about derivatives tofind that f0(x) = (x4+ 2x)0= (x4)0+ (2x)0= 4x3+ 2 · (x)0= 4x3+ 2. Alternatively, we can go thelong way (since this was in section 3.2) to find the derivative using the definition:f0(x) = limh→0f(x + h) − f(x)h= limh→0(x + h)4+ 2(x + h) − (x4+ 2x)h= limh→0(x2+ 2xh + h2)2+ 2x + 2h − x4− 2xh= limh→0x4+ 2x3h + x2h2+ 2x3h + 4x2h2+ 2xh3+ x2h2+ 2xh3+ h4+ 2h − x4h= limh→0h(2x3+ x2h + 2x3+ 4x2h + 2xh2+ x2h + 2xh2+ h3+ 2)h= limh→04x3+ 6x2h + 4x2h + h3+ 2 = 4x3+ 26. (3.3.59) Find the first and second derivative of f (x) =x21+2x.f0(x) =x21 + 2x0=(1 + 2x) · (x2)0− (x2) · (1 + 2x)0(1 + 2x)2=(1 + 2x)(2x) − x2(2)(1 + 2x)2=2x + 4x2− 2x2(1 + 2x)2=2x + 2x2(1 + 2x)2f00(x) =2x + 2x2(1 + 2x)20=(1 + 2x)2(2 + 4x) − (2x + 2x2)(1 + 4x + 4x2)0(1 + 2x)4=(1 + 4x + 4x2)(2 + 4x) − (2x + 2x2)(4 + 8x)(1 + 2x)4= · · · [expand and simplify] · · ·=2 + 4x(1 + 2x)4=2(1 + 2x)(1 + 2x)4=2(1 +
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