Question 21. Differentiate (2x2 – 3) sin x with respect to x.
Solution:
We have,
=> y = (2x2 – 3) sin x
On differentiating both sides, we get,
On using product rule we get,
=
=
=
Question 22. Differentiate
Solution:
We have,
=> y =
On differentiating both sides, we get,
On using product rule we get,
=
=
=
=
Question 23. Differentiate
Solution:
We have,
=> y =
On differentiating both sides, we get,
On using product rule we get,
=
=
=
=
=
=
Question 24. Differentiate
Solution:
We have,
=> y =
On differentiating both sides, we get,
On using product rule we get,
=
=
=
=
=
Question 25. Differentiate
Solution:
We have,
=> y =
On differentiating both sides, we get,
On using product rule we get,
=
=
=
=
=
=
=
Question 26. Differentiate (ax + b)n (cx + d)m with respect to x.
Solution:
We have,
=> y = (ax + b)n (cx + d)m
On differentiating both sides, we get,
On using product rule we get,
=
=
=
=
Question 27. Differentiate in two ways, using product rule and otherwise, the function (1 + 2 tan x) (5 + 4 cos x). Verify that the answer are the same.
Solution:
We have,
=> y = (1 + 2 tan x) (5 + 4 cos x)
On using product rule we get,
=
= 10 sec2 x + 8 cos x sec2 x − 4 sin x − 8 sin x tan x
=
=
=
=
= 10 sec2 x + 8 cos x − 4 sin x
By using an alternate method, we have,
=
=
On using chain rule, we get,
= 0 − 4 sin x + 10 sec2 x + 8 cos x
= 10 sec2 x + 8 cos x − 4 sin x
Hence proved.
Question 28. Differentiate each of the following functions by the product rule and the other method and verify that answer from both the methods is the same.
(i) (3x2 + 2)2
Solution:
We have,
=> y = (3x2 + 2)2
On using product rule we get,
=
= 12x (3x2 + 2)
= 36 x3 + 24x
By using an alternate method, we have,
On using chain rule, we get,
= 36 x3 + 0 + 24 x
= 36 x3 + 24x
Hence proved.
(ii) (x + 2)(x + 3)
Solution:
We have,
=> y = (x + 2)(x + 3)
On using product rule we get,
= (x+3)(1)+(x+2)(1)
= x + 3 + x + 2
= 2x + 5
By using an alternate method, we have,
On using chain rule, we get,
= 2x + 5
Hence proved.
(iii) (3 sec x − 4 cosec x) (−2 sin x + 5 cos x)
Solution:
We have,
=> y = (3 sec x − 4 cosec x) (−2 sin x + 5 cos x)
On using product rule we get,
= (−2 sin x + 5 cos x) (3 sec x tan x + 4 cot x cosec x)+ (3 sec x − 4 cosec x) (−2 cos x − 5 sin x)
= −6 sin x sec x tan x − 8 sin x cot x cosec x + 15 cos x sec x tan x + 20 cos x cot x cosec x − 6 sec x cos x − 15 sec x sin x + 8 cosec x cos x + 20 cosec x sin x
= −6 tan2 x − 8 cot x + 15 tan x + 20 cot2 x − 6 − 15 tan x + 8 cot x + 20
= − 6 − 6 tan2 x + 20 cot2 x + 20
= −6 (1 + tan2 x) + 20 (cot2 x + 1)
= −6 sec2 x + 20 cosec2 x
By using an alternate method, we have,
=
=
On using chain rule, we get,
= −6 sec2 x − (−20 cosec2 x)
= −6 sec2 x + 20 cosec2 x
Hence proved.