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Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 11 |

Exercise | 11.3 |

Category | RD Sharma Solutions |

**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^{−1}x

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. 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,

=

=

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

**Solution:**

We have,, −∞ < x < 0

On putting 2^{x} = tan θ, we get,

=

Now, −∞ < x < 0

=> 0 < 2^{x} < 1

=> 0 < θ < π/4

=> 0 < 2θ < π/2

So, y = 2θ

= 2 tan^{−1} (2^{x})

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 a^{x} = tan θ, we get,

=

Now, −∞ < x < 0

=> 0 < a^{x} < 1

=> 0 < θ < π/4

=> 0 < 2θ < π/2

So, y = 2θ

= 2 tan^{−1} (a^{x})

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. Differentiate****with 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 x^{n} = tan θ, we get,

=

Now, 0 < x < ∞

=> 0 < x^{n} < ∞

=> 0 < θ < π/2

=> 0 < 2θ < π

So, y = 2θ

= 2 tan^{–1} (x^{n})

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. Differentiate****with respect to x.**

**Solution:**

We have,

=

Differentiating with respect to x, we get,

= 0 +

=

**Question 26. Differentiate****with respect to x.**

**Solution:**

We have,

=

Differentiating with respect to x, we get,

=

=

**Question 27. Differentiate****with respect to x.**

**Solution:**

We have,

=

=

=

=

Differentiating with respect to x, we get,

= 0 + 1

= 1

**Question 28. Differentiate****with respect to x.**

**Solution:**

We have,

=

=

=

Differentiating with respect to x, we get,

= 0 +

=

**Question 29. Differentiate****with respect to x.**

**Solution:**

We have,

=

=

=

Differentiating with respect to x, we get,

=

=

=

**Question 30. Differentiate****with respect to x.**

**Solution:**

We have,

=

=

Differentiating with respect to x, we get,

=

=

**Question 31. Differentiate****with 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. 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} x

Now, 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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