Page 88 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 4] DERIVATIVES 79

d ðu þ uÞðv þ vÞ uv u v þ v u þ u v
uv ¼ lim ¼ lim
dx x!0 x x!0 x

v u u dv du
¼ lim u þ v þ v ¼ u þ v
x!0 x x x dx dx
where it is noted that v ! 0as x ! 0, since v is supposed differentiable and thus continuous.
dy dy du
4.13. If y ¼ f ðuÞ where u ¼ gðxÞ, prove that ¼ assuming that f and g are differentiable.
dx du dx
Let x be given an increment x 6¼ 0. Then as a consequence u and y take on increments u and y
respectively, where
y ¼ f ðu þ uÞ f ðuÞ; u ¼ gðx þ xÞ gðxÞ ð1Þ
Note that as x ! 0, y ! 0 and u ! 0.
y dy
so that ! 0as u ! 0 and
If u 6¼ 0, let us write ¼
u du
dy
u þ u
du
y ¼ ð2Þ
If u ¼ 0for values of x,then(1)shows that y ¼ 0for these values of x. For such cases, we
define ¼ 0.
It follows that in both cases, u 6¼ 0or u ¼ 0, (2) holds. Dividing (2)by x 6¼ 0 and taking the limit
as x ! 0, we have

dy y dy u u dy u u
¼ lim ¼ lim þ lim þ lim lim
dx x!0 x x!0 du x x ¼ du x!0 x x!0 x!0 x
dy du du dy du
du dx dx du dx
¼ þ 0 ¼ ð3Þ
d d
4.14. Given ðsin xÞ¼ cos x and ðcos xÞ¼ sin x, derive the formulas
dx dx
d 2 d 1 1
ðtan xÞ¼ sec x; ðsin
ðaÞ ðbÞ xÞ¼ p ffiffiffiffiffiffiffiffiffiffiffiffiffi
dx dx 1 x 2
d d
d d sin x cos x dx ðsin xÞ sin x dx ðcos xÞ
dx dx cos x cos x
ðaÞ ðtan xÞ¼ ¼ 2
1
¼ x
ðcos xÞðcos xÞ ðsin xÞð sin xÞ 2
cos x cos x
¼ 2 ¼ 2
(b)If y ¼ sin 1 x,then x ¼ sin y. Taking the derivative with respect to x,
dy dy 1 1 1
1 ¼ cos y or ¼ q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ p ffiffiffiffiffiffiffiffiffiffiffiffiffi
dx dx ¼ cos y 2 1 x 2
1 sin y
We have supposed here that the principal value =2 @ sin 1 x @ =2, is chosen so that cos y is
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
1 sin y rather than cos y ¼ 1 sin y.
positive, thus accounting for our writing cos y ¼
d log e du
a
4.15. Derive the formula ðlog uÞ¼ ða > 0; a 6¼ 1Þ, where u is a differentiable function of x.
a
dx u dx
Consider y ¼ f ðuÞ¼ log u.By definition,
a
dy f ðu þ uÞ f ðuÞ log ðu þ uÞ log u
a
a
¼ lim ¼ lim
du u!0 u u!0 u
u= u
1 u þ u 1 u
¼ lim log a ¼ lim log a 1 þ
u!0 u u u!0 u u