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DERIVATIVES 4.1. (a) Let f x 3 x , x 6 3. 3 x Evaluate f 0 2 from the de nition.
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  f 2 h f 2 1 5 h 1 6h 6 lim 5 lim lim 6 h!0 h!0 h 1 h h!0 h 1 h h!0 1 h h
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f 0 2 lim
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Note: By using rules of di erentiation we nd f x
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d d 3 x 3 x 3 x 3 x 1 3 x 1 6 dx dx 3 x 2 3 x 2 3 x 2
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at all points x where the derivative exists. Putting x 2, we nd f 0 2 6. Although such rules are often useful, one must be careful not to apply them indiscriminately (see Problem 4.5).
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(b) Let f x
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p 2x 1. Evaluate f 0 5 from the de nition.
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p f 5 h f 5 9 2h 3 lim h!0 h!0 h h p p 9 2h 9 2 1 9 2h 3 9 2h 3 lim p p lim p lim h!0 h 9 2h 3 h!0 h 9 2h 3 h!0 9 2h 3 3
f 0 5 lim
d d By using rules of di erentiation we nd f 0 x 2x 1 1=2 1 2x 1 1=2 2x 1 2 dx dx 1=2 0 1=2 1 2x 1 . Then f 5 9 3.
4.2. (a) Show directly from de nition that the derivative of f x x3 is 3x2 . d p 1 x p . (b) Show from de nition that dx 2 x
DERIVATIVES
[CHAP. 4
f x h f x 1 x h 3 x3 h h 1 x3 3x2 h 3xh2 h3 x3 3x2 3xh h2 h Then f 0 x lim f x h f x lim h!0 h p p x h x h
f x h f x 3x2 h
The result follows by multiplying numerator and denominator by
p p x h x and then letting h ! 0.
4.3. If f x has a derivative at x x0 , prove that f x must be continuous at x x0 .
f x0 h f x0 lim f x0 h f x0 lim Thus or
f x0 h f x0 h; h
h 6 0
Then
f x0 h f x0 lim h f 0 x0 0 0 h!0 h lim f x0 h f x0
since f 0 x0 exists by hypothesis.
lim f x0 h f x0 0
showing that f x is continuous at x x0 .
4.4. Let f x
x sin 1=x; x 6 0 . 0; x 0 (a) Is f x continuous at x 0 (b) Does f x have a derivative at x 0
f 0 h f 0 f h f 0 h sin 1=h 0 1 lim lim lim sin h!0 h!0 h!0 h h h h
(a) By Problem 3.22(b) of 3, f x is continuous at x 0. b f 0 0 lim
which does not exist. This example shows that even though a function is continuous at a point, it need not have a derivative at the point, i.e., the converse of the theorem in Problem 4.3 is not necessarily true. It is possible to construct a function which is continuous at every point of an interval but has a derivative nowhere.
& 4.5. Let f x
x2 sin 1=x; x 6 0 . 0; x 0
f h f 0 h2 sin 1=h 0 1 lim lim h sin 0 h!0 h!0 h h h
(a) Is f x di erentiable at x 0 (b) Is f 0 x continuous at x 0
a f 0 0 lim
by Problem 3.13, 3. Then f x has a derivative (is di erentiable) at x 0 and its value is 0. (b) From elementary calculus di erentiation rules, if x 6 0,       d 1 d 1 1 d 2 x2 sin sin x x2 sin f 0 x dx x dx x x dx      1 1 1 1 1 2 sin 2x cos 2x sin x2 cos x x x x x
CHAP. 4]
DERIVATIVES
  1 1 Since lim f 0 x lim cos 2x sin does not exist (because lim cos 1=x does not exist), f 0 x x!0 x!0 x!0 x x cannot be continuous at x 0 in spite of the fact that f 0 0 exists. This shows that we cannot calculate f 0 0 in this case by simply calculating f 0 x and putting x 0, as is frequently supposed in elementary calculus. It is only when the derivative of a function is continuous at a point that this procedure gives the right answer. This happens to be true for most functions arising in elementary calculus.
4.6. Present an ;  de nition of the derivative of f x at x x0 .
f x has a derivative f 0 x0 at x x0 if, given any  > 0, we can nd  > 0 such that    f x0 h f x0   when 0 < jhj <  f 0 x0  <    h
RIGHT- AND LEFT-HAND DERIVATIVES 4.7. Let f x jxj. (a) Calculate the right-hand derivatives of f x at x 0. (b) Calculate the lefthand derivative of f x at x 0. (c) Does f x have a derivative at x 0 (d) Illustrate the conclusions in (a), (b), and (c) from a graph.
0 f 0 lim h!0
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