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

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 2^x = tan θ, we get,

=

=

=

=

=

=

=

= 2θ

= 2 tan⁻¹ (2^x)

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.

Question 37 Differentiate the following with respect to x :

(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 6^x = tan θ, we get,

=

=

=

=

=

=

=

= 2θ

= 2 tan⁻¹ (6^x)

Differentiating with respect to x, we get,

=

=

Summary

This section likely covers more advanced and complex topics such as:

• Differentiation of complex composite functions
• Higher-order derivatives of implicit functions
• Applications of differentiation in optimization problems
• Differentiation involving hyperbolic functions
• Leibniz's theorem for nth order derivatives