# 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

^{−1}xDifferentiating 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. Differentiate****with 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. Differentiate****with 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,

=

=

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