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 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.
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. Differentiate with 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,
=
=