Page 87 - Schaum's Outline of Theory and Problems of Advanced Calculus

78 DERIVATIVES [CHAP. 4

2
4.9. Find an equation for the tangent line to y ¼ x at the point where (a) x ¼ 1=3; ðbÞ x ¼ 1.
(a)From Problem 4.8, f ðx 0 Þ¼ 2x 0 so that f ð1=3Þ¼ 2=3. Then the equation of the tangent line is
0
0
or y ¼ ðx Þ;
0 1 2 1 2 1
3
y f ðx 0 Þ¼ f ðx 0 Þðx x 0 Þ 9 3 3 i:e:; y ¼ x 9
(b)As in part (a), y f ð1Þ¼ f ð1Þðx 1Þ or y 1 ¼ 2ðx 1Þ, i.e., y ¼ 2x 1.
0
DIFFERENTIALS
3
4.10. If y ¼ f ðxÞ¼ x 6x, find (a) y; ðbÞ dy; ðcÞ y dy.
3 3
ðaÞ y ¼ f ðx þ xÞ f ðxÞ¼fðx þ xÞ 6ðx þ xÞg fx 6xg
2
2
3
3
3
¼ x þ 3x x þ 3xð xÞ þð xÞ 6x 6 x x þ 6x
2 2 3
¼ð3x 6Þ x þ 3xð xÞ þð xÞ
2
2
(b) dy ¼ principal part of y ¼ð3x 6Þ x ¼ð3x 6Þdx, since by definition x ¼ dx.
2
2
2
Note that f ðxÞ¼ 3x 6 and dy ¼ð3x 6Þdx, i.e., dy=dx ¼ 3x 6. It must be emphasized that
0
dy and dx are not necessarily small.
2
3
2
(c) From (a) and (b), y dy ¼ 3xð xÞ þð xÞ ¼ x, where ¼ 3x x þð xÞ .
y dy
Note that ! 0as x ! 0, i.e., ! 0as x ! 0. Hence y dy is an infinitesimal of
x
higher order than x (see Problem 4.83).
In case x is small, dy and y are approximately equal.
p ffiffiffiffiffi
3
4.11. Evaluate 25 approximately by use of differentials.
If x is small, y ¼ f ðx þ xÞ f ðxÞ¼ f ðxÞ x approximately.
0
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p ffiffiffi p ffiffiffi 1 2=3
3 x. Then 3 3 x x x (where denotes approximately equal to).
3
Let f ðxÞ¼ x þ x
If x ¼ 27 and x ¼ 2, we have
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffi p ffiffiffiffiffi
3 3 1 2=3 3
3
27 2 27 ð27Þ ð 2Þ; i.e., 25 3 2=27
ffiffiffiffiffi
p
Then 3 25 3 2=27 or 2.926.
3
If is interesting to observe that ð2:926Þ ¼ 25:05, so that the approximation is fairly good.
DIFFERENTIATION RULES: DIFFERENTIATION OF ELEMENTARY FUNCTIONS
d d d
4.12. Prove the formula f f ðxÞgðxÞg ¼ f ðxÞ gðxÞþ gðxÞ f ðxÞ, assuming f and g are differentiable.
dx dx dx
By definition,
d f ðx þ xÞgðx þ xÞ f ðxÞgðxÞ
f f ðxÞgðxÞg ¼ lim
dx x!0 x
¼ lim f ðx þ xÞfgðx þ xÞ gðxÞg þ gðxÞf f ðx þ xÞ f ðxÞg
x!0 x

gðx þ xÞ gðxÞ f ðx þ xÞ f ðxÞ
¼ lim f ðx þ xÞ þ lim gðxÞ
x!0 x x!0 x
d d
dx dx
¼ f ðxÞ gðxÞþ gðxÞ f ðxÞ
Another method:
Let u ¼ f ðxÞ, v ¼ gðxÞ. Then u ¼ f ðx þ xÞ f ðxÞ and v ¼ gðx þ xÞ gðxÞ, i.e., f ðx þ xÞ¼
u þ u, gðx þ xÞ¼ v þ v. Thus