# Class 11 RD Sharma Solutions – Chapter 30 Derivatives – Exercise 30.4 | Set 3

Last Updated : 16 May, 2021

### Question 21. Differentiate (2x2 â€“ 3) sin x with respect to x.

Solution:

We have,

=> y = (2x2 â€“ 3) sin x

On differentiating both sides, we get,

On using product rule we get,

### Question 22. Differentiate  with respect to x.

Solution:

We have,

=> y =

On differentiating both sides, we get,

On using product rule we get,

### Question 23. Differentiate  with respect to x.

Solution:

We have,

=> y =

On differentiating both sides, we get,

On using product rule we get,

### Question 24. Differentiate  with respect to x.

Solution:

We have,

=> y =

On differentiating both sides, we get,

On using product rule we get,

### Question 25. Differentiate  with respect to x.

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 âˆ’ 4 sin x âˆ’ 8 sin x tan x

=

= 10 sec2 x + 8 cos x âˆ’ 4 sin x

By using an alternate method, we have,

On using chain rule, we get,

= 0 âˆ’ 4 sin x + 10 sec2 x + 8 cos x

= 10 sec2 x + 8 cos x âˆ’ 4 sin x

Hence proved.

### (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 âˆ’ 4 cosec x) (âˆ’2 sin x + 5 cos x)

Solution:

We have,

=> y = (3 sec x âˆ’ 4 cosec x) (âˆ’2 sin x + 5 cos x)

On using product rule we get,

= (âˆ’2 sin x + 5 cos x) (3 sec x tan x + 4 cot x cosec x)+ (3 sec x âˆ’ 4 cosec x) (âˆ’2 cos x âˆ’ 5 sin x)

= âˆ’6 sin x sec x tan x âˆ’ 8 sin x cot x cosec x + 15 cos x sec x tan x  + 20 cos x cot x cosec x âˆ’ 6 sec x cos x âˆ’ 15 sec x sin x + 8 cosec x cos x + 20 cosec x sin x

= âˆ’6 tan2 x âˆ’ 8 cot x + 15 tan x + 20 cot2 x âˆ’ 6 âˆ’ 15 tan x + 8 cot x + 20

= âˆ’ 6 âˆ’ 6 tan2 x + 20 cot2 x + 20

= âˆ’6 (1 + tan2 x) + 20 (cot2 x + 1)

= âˆ’6 sec2 x + 20 cosec2 x

By using an alternate method, we have,

On using chain rule, we get,

= âˆ’6 sec2 x âˆ’ (âˆ’20 cosec2 x)

= âˆ’6 sec2 x + 20 cosec2 x

Hence proved.

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