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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 36
Total: 72
Find the derivative of the following functions from first principle.
(i) x3 – 27
(ii) (x – 1)(x – 2)
(iii)
1
x2
(iv)
x+1
x−1
Solution:  
(i) Let f (x) = x3 – 27
We have,
d
dx
 (f (x)) =
lim
h→0
f(x+h)−f(x)
h
=
lim
h→0
[(x+h)3−27]−[x3−27]
h
=
lim
h→0
[
x3+3x2h+3xh2+h3−27−x3+27
h
]
=
lim
h→0
3x2h+3xh2+h3
h
=
lim
h→0
[
3x2+3xh+h2
h
]
=
lim
h→0
(3x2+3xh+h2)
= 3x2 + 3x (0) + (0)2 = 3x2.
(ii) Let f(x) = (x – 1) (x – 2)
We have,
d
dx
 (f (x)) =
lim
h→0
f(x+h)−f(x)
h
=
lim
h→0
(x+h−1)(x+h−2)−(x−1)(x−2)
h

=
lim
h→0
[(x+h)2−2(x+h)−(x+h)+2]−[x2−2x−x+2]
h

=
lim
h→0
[x2+h2+2xh−3(x+h)+2]−[x2−3x+2]
h

=
lim
h→0
h2+2xh−3h
h
=
lim
h→0
 (h + 2x - 3) = 2x - 3
(iii) Let f (x) =
1
x2
We have,
d
dx
 (f (x)) =
lim
h→0
f(x+h)−f(x)
h
=
lim
h→0
[
1
(x+h)2
−
1
x2
h
] =
lim
h→0
[
x2−(x+h)2
(x+h)2x2h
] =
lim
h→0
[
x2−x2−h2−2xh
(x+h)2.x2.h
]
=
lim
h→0
[
−h2−2xh
(x+h)2.x2.h
] =
lim
h→0
(−
h
h
)[
h+2x
(x+h)2.x2
]
= -
lim
h→0
(
h+2x
(x+h)2.x2
) =
−2x
x2.x2
 = −
2
x3
(iv) Let f (x) =
x+1
x−1
We have,
d
dx
 (f (x)) =
lim
h→0
f(x+h)−f(x)
h
d
dx
 (f (x)) =
lim
h→0
[
(x+h+1)
(x+h−1)
−
(x+1)
(x−1)
h
]
=
lim
h→0
[
(x+h+1)(x−1)−(x+1)(x+h−1)
h(x+h−1)(x−1)
]

=
lim
h→0
[
x2+hx+x−x−h−1−[x2+hx−x+x+h−1]
h(x+h−1)(x−1)
]

=
lim
h→0
[
(x2+hx−h−1)−(x2+hx+h−1)
h(x+h−1)(x−1)
]

=
lim
h→0
[
x2+hx−h−1−x2−hx−h−1
h(x+h−1)(x−1)
]

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