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

Question 1. Differentiate, 1/√2 < x < 1 with respect to x.

Solution:

We have,

, 1/√2 < x < 1.

On putting x = cos θ, we get,

y =

=

= cos⁻¹(2cos θ sin θ)

= cos⁻¹(sin 2θ)

=

Now, 1/√2 < x < 1

=> 1/√2 < cos θ < 1

=> 0 < θ < π/4

=> 0 < 2θ < π/2

=> 0 > −2θ > −π/2

=> π/2 > (π/2−2θ) > 0

So, y =

Differentiating with respect to x, we get,

=

=

Question 2. Differentiate,−1 < x < 1 with respect to x.

Solution:

We have,,−1 < x < 1.

On putting x = cos 2θ, we get,

y =

=

=

=

Now, −1 < x < 1

=> −1 < cos 2θ < 1

=> 0 < 2θ < π

=> 0 < θ < π/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 2θ, we get,

y =

=

=

Now, 0 < x < 1

=> 0 < cos 2θ < 1

=> 0 < 2θ < π/2

=> 0 < θ < π/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 θ, we get,

y =

=

=

Now, 0 < x < 1

=> 0 < cos θ < 1

=> 0 < θ < π/2

So, y = cos⁻¹x

Differentiating with respect to x, we get,

=

Question 5. Differentiate, −a < x < a with respect to x.

Solution:

We have,, −a < x < a

On putting x = a sin θ, we get,

y =

=

=

=

Now, −a < x < a

=> −1 < x/a < 1

=> −π/2 < θ < π/2

So,

Differentiating with respect to x, we get,

=

=

=

Question 6. Differentiatewith respect to x.

Solution:

We have,

On putting x = a tan θ, we get,

y =

=

=

=

= θ

=

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 θ, we get,

y =

=

=

Now, 0 < x < 1

=> 0 < cos θ < 1

=> 0 < θ < π/2

=> 0 < 2θ < π

=> π/2 > (π/2−2θ) > −π/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 θ, we get,

y =

=

=

Now, 0 < x < 1

=> 0 < sin θ < 1

=> 0 < θ < π/2

=> 0 < 2θ < π

=> π/2 > (π/2−2θ) > −π/2

So, y =

Differentiating with respect to x, we get,

=

=

Question 9. Differentiatewith respect to x.

Solution:

We have,

Putting x = cot θ, we get,

y =

=

=

=

= θ

=

Differentiating with respect to x, we get,

=

=

=

Question 10. Differentiate, −3π/4 < x < π/4 with respect to x.

Solution:

We have,, −3π/4 < x < π/4

=

=

Now, −3π/4 < x < π/4

=> −π/2 < (x+π/4) < π/2

So, y =

Differentiating with respect to x, we get,

= 1 + 0

= 1

Question 11. Differentiate, −π/4 < x < π/4 with respect to x.

Solution:

We have,, −π/4 < x < π/4

=

=

Now, −π/4 < x < π/4

=> −π/2 < (x−π/4) < 0

So, y =

=

Differentiating with respect to x, we get,

= −1 + 0

= −1

Question 12. Differentiate, −1 < x < 1 with respect to x.

Solution:

We have,, −1 < x < 1

On putting x = sin θ, we get,

y =

=

=

=

Now, −1 < x < 1

=> −1 < sin θ < 1

=> −π/2 < θ < π/2

=> −π/4 < θ/2 < π/4

So, y =

Differentiating with respect to x, we get,

=

Question 13. Differentiate, −a < x < a with respect to x.

Solution:

We have,, −a < x < a

On putting x = a sin θ, we get,

=

=

=

=

Now, −a < x < a

=> −1 < x/a < 1

=> −π/2 < θ < π/2

=> −π/4 < θ/2 < π/4

So, y =

Differentiating with respect to x, we get,

=

=

=

Question 14. Differentiate, −1 < x < 1 with respect to x.

Solution:

We have,, −1 < x < 1

On putting x = sin θ, we get,

=

=

Now, −1 < x < 1

=> −1 < sin θ < 1

=> −π/2 < θ < π/2

=> −π/2 < (θ+π/4) < 3π/4

So, y =

Differentiating with respect to x, we get,

=

=

Question 15. Differentiate, −1 < x < 1 with respect to x.

Solution:

We have,, −1 < x < 1

On putting x = sin θ, we get,

=

=

Now, −1 < x < 1

=> −1 < sin θ < 1

=> −π/2 < θ < π/2

=> −3π/4 < (θ−π/4) < π/4

So, y =

=

Differentiating with respect to x, we get,

=

=

Question 16. Differentiate, −1/2 < x < 1/2 with respect to x.

Solution:

We have,, −1/2 < x < 1/2

On putting 2x = tan θ, we get,

=

Now, −1/2 < x < 1/2

=> −1 < 2x < 1

=> −1 < tan θ < 1

=> −π/4 < θ < π/4

=> −π/2 < 2θ < π/2

Therefore, y = 2 tan⁻¹ (2x)

Differentiating with respect to x, we get,

=

=