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

**Question 33. Differentiate****with respect to x.**

**Solution:**

We have,

=

Differentiating with respect to x, we get,

=

=

=

**Question 34. Differentiate****with 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}xNow, 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}xNow, L.H.S. =

=

= R.H.S.

Hence proved.

**Question 37 Differentiate the following with respect to x :**

**(i) cos**^{−1} (sin x)

^{−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}xNow, 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 + a^{2} = b.

^{−1}(a + bx) takes the value of 1 at x = 0, prove that 1 + a

^{2}= b.

**Solution:**

We have, y = tan

^{−1}(a + bx)Differentiating with respect to x, we get,

=

At x = 0, we have,

=>= 1

=>= 1

=> 1 + a

^{2}= 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. Differentiate****with 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,

=

=

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