# Class 12 RD Sharma Solutions – Chapter 11 Differentiation – Exercise 11.3 | Set 3

Last Updated : 08 May, 2021

### 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 Î¸, we get,

=

=

=

=

=

=

=

= 2Î¸

= 2 tanâˆ’1 (2x)

Differentiating with respect to x, we get,

=

=

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

Solution:

We have,

=

On putting x = tan Î¸, we get,

y =

=

=

=

=

=

Now, 0 < x < 1

=> 0 < tan Î¸ < 1

=> 0 < Î¸ < Ï€/4

=> 0 < 2Î¸ < Ï€/2

So, y = 2Î¸ + 2Î¸

= 4Î¸

= 4 tanâˆ’1 x

Now, L.H.S. =

=

= R.H.S.

Hence proved.

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

Solution:

We have,

On putting x = tan Î¸, we get,

=

=

=

=

Now, 0 < x < âˆž

=> 0 < tan Î¸ < âˆž

=> 0 < Î¸ < Ï€/2

So, y = Î¸ + Î¸

= 2Î¸

= 2 tanâˆ’1 x

Now, L.H.S. =

=

= R.H.S.

Hence proved.

### (i) cosâˆ’1 (sin x)

Solution:

We have, y = cosâˆ’1 (sin x)

=

=

Differentiating with respect to x, we get,

= 0 âˆ’ 1

= âˆ’1

### (ii)

Solution:

We have, y =

On putting x = tan Î¸, we get,

=

=

=

=

=

=

Differentiating with respect to x, we get,

= 0 +

=

### Question 38. Differentiate, 0 < x < Ï€/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 Î¸, we get,

y =

=

=

=

=

=

=

=

=

Here, 0 < x < âˆž

=> 0 < tan Î¸ < âˆž

=> 0 < Î¸ < Ï€/2

=> 0 < 2Î¸ < Ï€

So, y = 2Î¸ + 2Î¸

= 4Î¸

= 4 tanâˆ’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 2Î¸, we get,

=

=

=

=

=

=

=

=

Differentiating with respect to x, we get,

=

=

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

Solution:

We have,

On putting 2x = cos Î¸, we get,

=

=

Now, 0 < x < 1/2

=> 0 < 2x < 1

=> 0 < cos Î¸ < 1

=> 0 < Î¸ < Ï€/2

and 0 > âˆ’Î¸ > âˆ’Ï€/2

=> Ï€/2 > (Ï€/2 âˆ’Î¸) > 0

So, y =

= Ï€ âˆ’ Î¸

= Ï€ âˆ’ cosâˆ’1 (2x)

Differentiating with respect to x, we get,

=

=

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

Solution:

We have, y = tanâˆ’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, âˆ’1/2 < x < 0, find.

Solution:

We have,

On putting 2x = cos Î¸, we get,

=

=

Now, âˆ’1/2 < x < 0

=> âˆ’1 < 2x < 0

=> âˆ’1 < cos Î¸ < 0

=> Ï€/2 < Î¸ < Ï€

and âˆ’Ï€/2 > âˆ’Î¸ > âˆ’Ï€

=> 0 > (Ï€/2 âˆ’Î¸) > âˆ’Ï€/2

So, y =

= âˆ’Ï€ + 3Î¸

= âˆ’Ï€ + 3 cosâˆ’1 (2x)

Differentiating with respect to x, we get,

= 0 +

=

### Question 45. If, find.

Solution:

We have,

On putting x = cos 2Î¸, we get,

=

=

=

=

=

=

=

=

Differentiating with respect to x, we get,

= 0 âˆ’

=

### Question 46. If , find.

Solution:

We have,

On putting x = cos Î¸, we get,

=

=

Let

=> sin Ã˜ =

=> sin Ã˜ =

=> sin Ã˜ =

=> sin Ã˜ =

=> sin Ã˜ =

So, y =

=

= Ã˜ + Î¸

=

Differentiating with respect to x, we get,

= 0 +

=

### Question 47. Differentiatewith respect to x.

Solution:

We have,

=

=

On putting 6x = tan Î¸, we get,

=

=

=

=

=

=

=

= 2Î¸

= 2 tanâˆ’1 (6x)

Differentiating with respect to x, we get,

=

=

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