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

**Question 21. Differentiate (2x**^{2} – 3) sin x with respect to x.

^{2}– 3) sin x with respect to x.

**Solution:**

We have,

=> y = (2x

^{2 }– 3) sin xOn 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.

^{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 sec

^{2}x + 8 cos x sec^{2}x − 4 sin x − 8 sin x tan x=

=

=

=

= 10 sec

^{2}x + 8 cos x − 4 sin xBy using an alternate method, we have,

=

=

On using chain rule, we get,

= 0 − 4 sin x + 10 sec

^{2}x + 8 cos x= 10 sec

^{2}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) (3x**^{2} + 2)^{2}

^{2}+ 2)

^{2}

**Solution:**

We have,

=> y = (3x

^{2}+ 2)^{2}On using product rule we get,

=

= 12x (3x

^{2}+ 2)= 36 x

^{3}+ 24xBy using an alternate method, we have,

On using chain rule, we get,

= 36 x

^{3}+ 0 + 24 x= 36 x

^{3}+ 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 tan

^{2}x − 8 cot x + 15 tan x + 20 cot^{2}x − 6 − 15 tan x + 8 cot x + 20= − 6 − 6 tan

^{2}x + 20 cot^{2}x + 20= −6 (1 + tan

^{2}x) + 20 (cot^{2}x + 1)= −6 sec

^{2}x + 20 cosec^{2}xBy using an alternate method, we have,

=

=

On using chain rule, we get,

= −6 sec

^{2}x − (−20 cosec^{2}x)= −6 sec

^{2}x + 20 cosec^{2}x

Hence proved.

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