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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 43
Total: 72
Find the derivative of the following functions :
(i) sin x cos x
(ii) sec x
(iii) 5 sec x + 4 cos x
(iv) cosec x
(v) 3 cot x + 5 cosec x
(vi) 5 sin x – 6 cos x + 7
(vii) 2 tan x – 7 sec x.
Solution:  
(i) Let f(x) = sin x cos x ... (1)
Differentiating (1) with respect to x, we get
f ′(x) = (sin x)′ cos x + (cos x)′ sin x = cosx · cosx + (– sin x) · sin x
= cos2x–sin2x
∴ f ′(x) = cos 2x.
(ii) Let f(x) = sec x
⇒ f (x) =
1
cosx
 ⇒ f (x) = (cosx)−1 ... (1)
Differentiating (1) with respect to x, we get f ′(x) = – (cosx)–2 · (– sinx)
⇒ f' (x) =
sinx
cos2x
 ⇒ f' (x) =
sinx
cosx
.
1
cosx
∴ f ′(x) = tanx · secx.
(iii) Let f (x) = 5 secx + 4 cosx
⇒ f (x) =
5
cosx
+4cosx ... (1)
Differentiating (1) with respect to x, we get
f ′(x) = 5(–1) (cosx)–2 (– sin x) + 4 (– sin x)
=
5sinx
cos2x
 - 4 sin x =
5sinx
cosx
.
1
cosx
 - 4 sin x
∴ f' (x) = 5 tan x . sec x - 4 sin x.
(iv) Let f(x) = cosec x
⇒ f (x) =
1
sinx
 ⇒ f (x) = (sinx)−1 ... (1)
Differentiating (1) with respect to x, we get
f ′(x) = (– 1) (sinx)–2 · cos x =
−cosx
sin2x
 =
−cosx
sinx
.
1
sinx
∴ f ′(x) = – cot x cosec x
(v) Let f(x) = 3 cot x + 5 cosec x
⇒ f (x) =
3cosx
sinx
+
5
sinx
 ... (1)
Differentiating (1) with respect to x, we get
f' (x) = 3[
sinx(cosx)′−cosx.(sinx)′
sin2x
] + 5[
sinx(1)′−1.(sinx)′
sin2x
]
= 3[
sinx.(−sinx)−cosx(cosx)
sin2x
] + 5[
0−cosx
sin2x
]
= 3[
−sin2x−cos2x
sin2x
] + 5[
−cosx
sin2x
]
= −3[
sin2x+cos2x
sin2x
] - 5[
cosx
sinx
.
1
sinx
]
= −3[
1
sin2x
] - 5 [cot x . cosec x]
= – 3 cosec2 x – 5 cot x · cosec x = – cosec x [3 cosec x + 5 cot x].
(vi) Let f(x) = 5 sin x – 6 cos x + 7 ... (1)
Differentiating (1) with respect to x, we get
f ′(x) = 5 cos x – 6 (– sin x) + 0
∴ f ′(x) = 5 cos x + 6 sin x.
(vii) Let f(x) = 2 tan x – 7 sec x ... (1)
Differentiating (1) with respect to x, we get
f ′(x) = 2 sec2 x – 7 sec x tan x
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