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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 44
Total: 72
Find the derivative of the following functions from first principle:
(i) –x
(ii) (–x)–1
(iii) sin (x + 1)
(iv) cos (x−
π
8
)
Solution:  
(i) Let f(x) = – x
We have
d
dx
 (f (x)) =
lim
h→0
f(x+h)−f(x)
h
=
lim
h→0
−(x+h)−(−x)
h
=
lim
h→0
−x−h+x
h
=
lim
h→0
−h
h
=
lim
h→0
(−1) = - 1
(ii) Let f(x) = (–x)–1
⇒ f (x) = −
1
x
We have,
d
dx
 (f (x)) =
lim
h→0
[
f(x+h)−f(x)
h
]
=
lim
h→0
[−1/x+h−(−
1
x
)]
h
 =
lim
h→0
[
−1
x+h
+
1
x
]
h
=
lim
h→0
[
−x+x+h
x(x+h)h
]
=
lim
h→0
[
h
hx(x+h)
] =
1
x2
(iii) Let f(x) = sin (x + 1)
We have,
d
dx
 (f (x)) =
lim
h→0
[f(x+h)−f(x)]
h
=
lim
h→0
sin(x+h+x)−sin(x+1)
h
=
lim
h→0
2cos(
x+h+1+x+1
2
)sin(
x+h+1−x−1
2
)
h

=
lim
h→0
2[cos(
2(x+1)+h
2
)sin
h
2
]
2×(
h
2
)

lim
h→0
cos(x+1+
h
2
)(
sin
h
2
h
2
)

=
[
lim
h→0
cos(x+1+
h
2
)][
lim
h→0
sin
h
2
h
2
]

= cos (x + 1) × (1) = cos (x + 1).
(iv) Let f (x) = cos (x−
π
8
)
We have,
d
dx
 (f (x)) =
lim
h→0
cos(x+h−
π
8
)−cos(x−
π
8
)
h

=
lim
h→0
[
−2sin(
x+h−
π
8
+x−
π
8
2
)sin(
s+h−
π
8
+x−
π
8
2
)
h
]

=
lim
h→0
[
−2sin(x−
π
8
+
h
2
).sin(
h
2
)
h
]

=
[
lim
h→0
sin(x−
π
8
+
h
2
)][
lim
h→0
sin(
h
2
)
h
2
]

= - sin (x−
π
8
) (1)
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