# RD Sharma Class 12 Ex 11.3 Solutions Chapter 11 Differentiation

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## RD Sharma Class 12 Ex 11.3 Solutions Chapter 11 Differentiation

### 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,

=

=

### Question 17. Differentiate, −∞ < x < 0 with respect to x.

Solution:

We have,, −∞ < x < 0

On putting 2x = tan θ, we get,

=

Now, −∞ < x < 0

=> 0 < 2x < 1

=> 0 < θ < π/4

=> 0 < 2θ < π/2

So, y = 2θ

= 2 tan−1 (2x)

Differentiating with respect to x, we get,

=

=

### Question 18. Differentiate, a > 1, −∞ < x < 0 with respect to x.

Solution:

We have,, −∞ < x < 0

On putting ax = tan θ, we get,

=

Now, −∞ < x < 0

=> 0 < ax < 1

=> 0 < θ < π/4

=> 0 < 2θ < π/2

So, y = 2θ

= 2 tan−1 (ax)

Differentiating with respect to x, we get,

=

=

### Question 19. Differentiate, 0 < x < 1 with respect to x.

Solution:

We have,, 0 < x < 1

On putting x = cos 2θ, we get,

=

=

=

=

Now, 0 < x < 1

=> 0 < cos 2θ < 1

=> 0 < 2θ < π/2

=> 0 < θ < π/4

=> π/4 < (θ+π/4) < π/2

So, y =

=

Differentiating with respect to x, we get,

=

=

### Question 20. Differentiate, x ≠ 0 with respect to x.

Solution:

We have,

On putting ax = tan θ, we get,

=

=

=

=

=

=

Differentiating with respect to x, we get,

=

### Question 21. Differentiate, −π < x < π with respect to x.

Solution:

We have,, −π < x < π

=

=

=

Differentiating with respect to x, we get,

=

### Question 22. Differentiatewith respect to x.

Solution:

We have,

On putting x = cot θ, we get,

=

=

= θ

= cot−1 x

Differentiating with respect to x, we get,

=

### Question 23. Differentiate, 0 < x < ∞ with respect to x.

Solution:

We have,,0 < x < ∞

On putting xn = tan θ, we get,

=

Now, 0 < x < ∞

=> 0 < xn < ∞

=> 0 < θ < π/2

=> 0 < 2θ < π

So, y = 2θ

= 2 tan–1 (xn)

Differentiating with respect to x, we get,

=

=

### Question 24. Differentiate, x ∈ R with respect to x.

Solution:

We have,

=

=

Differentiating with respect to x, we get,

= 0

### Question 25. Differentiatewith respect to x.

Solution:

We have,

=

Differentiating with respect to x, we get,

= 0 +

=

### Question 26. Differentiatewith respect to x.

Solution:

We have,

=

Differentiating with respect to x, we get,

=

=

### Question 27. Differentiatewith respect to x.

Solution:

We have,

=

=

=

=

Differentiating with respect to x, we get,

= 0 + 1

= 1

### Question 28. Differentiatewith respect to x.

Solution:

We have,

=

=

=

Differentiating with respect to x, we get,

= 0 +

=

### Question 29. Differentiatewith respect to x.

Solution:

We have,

=

=

=

Differentiating with respect to x, we get,

=

=

=

### Question 30. Differentiatewith respect to x.

Solution:

We have,

=

=

Differentiating with respect to x, we get,

=

=

### Question 31. Differentiatewith respect to x.

Solution:

We have,

=

=

Differentiating with respect to x, we get,

=

=

### Question 32. Differentiate, −π/4 < x < π/4 with respect to x.

Solution:

We have,, −π/4 < x < π/4

=

=

=

=

=

Differentiating with respect to x, we get,

= 0 + 1

= 1

### Question 33. Differentiatewith respect to x.

Solution:

We have,

=

Differentiating with respect to x, we get,

=

=

=

### Question 34. Differentiatewith respect to x.

Solution:

We have,

On putting 2x = tan θ, we get,

=

=

=

=

=

=

=

= 2θ

= 2 tan−1 (2x)

Differentiating with respect to x, we get,

=

=

### Question 35. If, 0 < x < 1, prove that.

Solution:

We have,

=

On putting x = tan θ, we get,

y =

=

=

=

=

=

Now, 0 < x < 1

=> 0 < tan θ < 1

=> 0 < θ < π/4

=> 0 < 2θ < π/2

So, y = 2θ + 2θ

= 4θ

= 4 tan−1 x

Now, L.H.S. =

=

= R.H.S.

Hence proved.

### Question 36. If, 0 < x < ∞, prove that.

Solution:

We have,

On putting x = tan θ, we get,

=

=

=

=

Now, 0 < x < ∞

=> 0 < tan θ < ∞

=> 0 < θ < π/2

So, y = θ + θ

= 2θ

= 2 tan−1 x

Now, L.H.S. =

=

= R.H.S.

Hence proved.

### (i) cos−1 (sin x)

Solution:

We have, y = cos−1 (sin x)

=

=

Differentiating with respect to x, we get,

= 0 − 1

= −1

### (ii)

Solution:

We have, y =

On putting x = tan θ, we get,

=

=

=

=

=

=

Differentiating with respect to x, we get,

= 0 +

=

### Question 38. Differentiate, 0 < x < π/2 with respect to x.

Solution:

We have,

=

=

=

=

=

=

=

=

Differentiating with respect to x, we get,

=

### Question 39. If, x > 0, prove that.

Solution:

We have,

=

On putting x = tan θ, we get,

y =

=

=

=

=

=

=

=

=

Here, 0 < x < ∞

=> 0 < tan θ < ∞

=> 0 < θ < π/2

=> 0 < 2θ < π

So, y = 2θ + 2θ

= 4θ

= 4 tan−1 x

Now, L.H.S. =

=

= R.H.S.

Hence proved.

### Question 40. If, x > 0, find.

Solution:

We have,

=

=

Differentiating with respect to x, we get,

= 0

### Question 41. If, find.

Solution:

We have,

On putting x = cos 2θ, we get,

=

=

=

=

=

=

=

=

Differentiating with respect to x, we get,

=

=

### Question 42. If , 0 < x < 1/2, find.

Solution:

We have,

On putting 2x = cos θ, we get,

=

=

Now, 0 < x < 1/2

=> 0 < 2x < 1

=> 0 < cos θ < 1

=> 0 < θ < π/2

and 0 > −θ > −π/2

=> π/2 > (π/2 −θ) > 0

So, y =

= π − θ

= π − cos−1 (2x)

Differentiating with respect to x, we get,

=

=

### Question 43. If the derivative of tan−1 (a + bx) takes the value of 1 at x = 0, prove that 1 + a2 = b.

Solution:

We have, y = tan−1 (a + bx)

Differentiating with respect to x, we get,

=

At x = 0, we have,

=>= 1

=>= 1

=> 1 + a2 = b

Hence proved.

### Question 44. If, −1/2 < x < 0, find.

Solution:

We have,

On putting 2x = cos θ, we get,

=

=

Now, −1/2 < x < 0

=> −1 < 2x < 0

=> −1 < cos θ < 0

=> π/2 < θ < π

and −π/2 > −θ > −π

=> 0 > (π/2 −θ) > −π/2

So, y =

= −π + 3θ

= −π + 3 cos−1 (2x)

Differentiating with respect to x, we get,

= 0 +

=

### Question 45. If, find.

Solution:

We have,

On putting x = cos 2θ, we get,

=

=

=

=

=

=

=

=

Differentiating with respect to x, we get,

= 0 −

=

### Question 46. If , find.

Solution:

We have,

On putting x = cos θ, we get,

=

=

Let

=> sin Ø =

=> sin Ø =

=> sin Ø =

=> sin Ø =

=> sin Ø =

So, y =

=

= Ø + θ

=

Differentiating with respect to x, we get,

= 0 +

=

### Question 47. Differentiatewith respect to x.

Solution:

We have,

=

=

On putting 6x = tan θ, we get,

=

=

=

=

=

=

=

= 2θ

= 2 tan−1 (6x)

Differentiating with respect to x, we get,

=

=

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