unit 1 differential calculus - IGNOU

(ii)
(iii)
d dg ( x)
df ( x)
( f ( x)
. g ( x))
= f ( x)
+ g ( x)
dx
dx
dx
d
dx
df ( x)
dg(
x)
⎛
g ( x)
f ( x)
f ( x)
⎞
−
dx
dx
⎜
g ( x)
⎟ =
2
⎝ ⎠
[ g ( x)]
provided g (x) ≠ 0.
We shall give the proof of the result (ii) below. Other results we leave as an
exercise for you.
Proof
Let y = f (x) . g (x)
Then
dy
dx
= lim
h→
0
f ( x + h)
g ( x + h)
− f ( x)
g ( x)
h
Now f ( x + h)
g ( x + h)
− f ( x)
g ( x)
∴
That is,
dy
dx
dy
dx
dy
dx
= f ( x + h)
[ g ( x + h)
− g ( x)]
+ g ( x)
[ f ( x + h)
− f ( x)]
⎧ [ g ( x + h)
− g ( x)]
[ f ( x + h)
− f ( x)]
⎫
⎨ f ( x + h)
+ g ( x)
⎬
⎩
h
⎭
= lim
h→
0
h
g ( x + h)
− g ( x)
= lim f ( x + h)
lim
h→
0
h →0
h
f ( x + h)
− f ( x)
+ g ( x)
. lim
h →0
h
dg ( x)
df ( x)
= f ( x)
+ g ( x)
,
dx dx
Use of the product rule (ii) and the result which states that ‘derivative of a
constant is zero’ gives
d
dx
where c is a constant.
Example 1.24
Solution
d
= [ cf ( x)]
= c f ( x)
,
dx
Find the derivative of y = 5x 2 sin x.
dy
dx
d
d
= 5 [ x (sin x)
+ sin x ( x
dx
dx
2 2
d
We derive (sin x)
as under :
dx
d
dx
(sin x)
= lim
h→
0
sin ( x + h)
− sin x
h
)]
Differential Calculus
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