# Class 12 RD Sharma Solutions – Chapter 11 Differentiation – Exercise 11.3 | Set 1

### Question 1. Differentiate, 1/âˆš2 < x < 1 with respect to x.

Solution:

We have,

, 1/âˆš2 < x < 1.

On putting x = cos Î¸, we get,

y =

=

= cosâˆ’1(2cos Î¸ sin Î¸)

= cosâˆ’1(sin 2Î¸)

=

Now, 1/âˆš2 < x < 1

=> 1/âˆš2 < cos Î¸ < 1

=> 0 < Î¸ < Ï€/4

=> 0 < 2Î¸ < Ï€/2

=> 0 > âˆ’2Î¸ > âˆ’Ï€/2

=> Ï€/2 > (Ï€/2âˆ’2Î¸) > 0

So, y =

Differentiating with respect to x, we get,

=

=

### Question 2. Differentiate,âˆ’1 < x < 1 with respect to x.

Solution:

We have,,âˆ’1 < x < 1.

On putting x = cos 2Î¸, we get,

y =

=

=

=

Now, âˆ’1 < x < 1

=> âˆ’1 < cos 2Î¸ < 1

=> 0 < 2Î¸ < Ï€

=> 0 < Î¸ < Ï€/2

So, y =

Differentiating with respect to x, we get,

=

### Question 3. Differentiate, 0 < x < 1 with respect to x.

Solution:

We have,, 0 < x < 1.

On putting x = cos 2Î¸, we get,

y =

=

=

Now, 0 < x < 1

=> 0 < cos 2Î¸ < 1

=> 0 < 2Î¸ < Ï€/2

=> 0 < Î¸ < Ï€/4

So,

Differentiating with respect to x, we get,

=

### Question 4. Differentiate, 0 < x < 1 with respect to x.

Solution:

We have,, 0 < x < 1

On putting x = cos Î¸, we get,

y =

=

=

Now, 0 < x < 1

=> 0 < cos Î¸ < 1

=> 0 < Î¸ < Ï€/2

So, y = cosâˆ’1x

Differentiating with respect to x, we get,

=

### Question 5. Differentiate, âˆ’a < x < a with respect to x.

Solution:

We have,, âˆ’a < x < a

On putting x = a sin Î¸, we get,

y =

=

=

=

Now, âˆ’a < x < a

=> âˆ’1 < x/a < 1

=> âˆ’Ï€/2 < Î¸ < Ï€/2

So,

Differentiating with respect to x, we get,

=

=

=

### Question 6. Differentiatewith respect to x.

Solution:

We have,

On putting x = a tan Î¸, we get,

y =

=

=

=

= Î¸

=

Differentiating with respect to x, we get,

=

=

=

### Question 7. Differentiate, 0 < x < 1 with respect to x.

Solution:

We have,, 0 < x < 1

On putting x = cos Î¸, we get,

y =

=

=

Now, 0 < x < 1

=> 0 < cos Î¸ < 1

=> 0 < Î¸ < Ï€/2

=> 0 < 2Î¸ < Ï€

=> Ï€/2 > (Ï€/2âˆ’2Î¸) > âˆ’Ï€/2

So, y =

Differentiating with respect to x, we get,

=

=

### Question 8. Differentiate, 0 < x < 1 with respect to x.

Solution:

We have, 0 < x < 1

On putting x = sin Î¸, we get,

y =

=

=

Now, 0 < x < 1

=> 0 < sin Î¸ < 1

=> 0 < Î¸ < Ï€/2

=> 0 < 2Î¸ < Ï€

=> Ï€/2 > (Ï€/2âˆ’2Î¸) > âˆ’Ï€/2

So, y =

Differentiating with respect to x, we get,

=

=

### Question 9. Differentiatewith respect to x.

Solution:

We have,

Putting x = cot Î¸, we get,

y =

=

=

=

= Î¸

=

Differentiating with respect to x, we get,

=

=

=

### Question 10. Differentiate, âˆ’3Ï€/4 < x < Ï€/4 with respect to x.

Solution:

We have,, âˆ’3Ï€/4 < x < Ï€/4

=

=

Now, âˆ’3Ï€/4 < x < Ï€/4

=> âˆ’Ï€/2 < (x+Ï€/4) < Ï€/2

So, y =

Differentiating with respect to x, we get,

= 1 + 0

= 1

### Question 11. Differentiate, âˆ’Ï€/4 < x < Ï€/4 with respect to x.

Solution:

We have,, âˆ’Ï€/4 < x < Ï€/4

=

=

Now, âˆ’Ï€/4 < x < Ï€/4

=> âˆ’Ï€/2 < (xâˆ’Ï€/4) < 0

So, y =

=

Differentiating with respect to x, we get,

= âˆ’1 + 0

= âˆ’1

### Question 12. Differentiate, âˆ’1 < x < 1 with respect to x.

Solution:

We have,, âˆ’1 < x < 1

On putting x = sin Î¸, we get,

y =

=

=

=

Now, âˆ’1 < x < 1

=> âˆ’1 < sin Î¸ < 1

=> âˆ’Ï€/2 < Î¸ < Ï€/2

=> âˆ’Ï€/4 < Î¸/2 < Ï€/4

So, y =

Differentiating with respect to x, we get,

=

### Question 13. Differentiate, âˆ’a < x < a with respect to x.

Solution:

We have,, âˆ’a < x < a

On putting x = a sin Î¸, we get,

=

=

=

=

Now, âˆ’a < x < a

=> âˆ’1 < x/a < 1

=> âˆ’Ï€/2 < Î¸ < Ï€/2

=> âˆ’Ï€/4 < Î¸/2 < Ï€/4

So, y =

Differentiating with respect to x, we get,

=

=

=

### Question 14. Differentiate, âˆ’1 < x < 1 with respect to x.

Solution:

We have,, âˆ’1 < x < 1

On putting x = sin Î¸, we get,

=

=

Now, âˆ’1 < x < 1

=> âˆ’1 < sin Î¸ < 1

=> âˆ’Ï€/2 < Î¸ < Ï€/2

=> âˆ’Ï€/2 < (Î¸+Ï€/4) < 3Ï€/4

So, y =

Differentiating with respect to x, we get,

=

=

### Question 15. Differentiate, âˆ’1 < x < 1 with respect to x.

Solution:

We have,, âˆ’1 < x < 1

On putting x = sin Î¸, we get,

=

=

Now, âˆ’1 < x < 1

=> âˆ’1 < sin Î¸ < 1

=> âˆ’Ï€/2 < Î¸ < Ï€/2

=> âˆ’3Ï€/4 < (Î¸âˆ’Ï€/4) < Ï€/4

So, y =

=

Differentiating with respect to x, we get,

=

=

### Question 16. Differentiate, âˆ’1/2 < x < 1/2 with respect to x.

Solution:

We have,, âˆ’1/2 < x < 1/2

On putting 2x = tan Î¸, we get,

=

Now, âˆ’1/2 < x < 1/2

=> âˆ’1 < 2x < 1

=> âˆ’1 < tan Î¸ < 1

=> âˆ’Ï€/4 < Î¸ < Ï€/4

=> âˆ’Ï€/2 < 2Î¸ < Ï€/2

Therefore, y = 2 tanâˆ’1 (2x)

Differentiating with respect to x, we get,

=

=

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