Differentiate
with respect to
if –1<x<1, x ≠ 0. (CBSE 2014,2016)
Let
and
.
We need to differentiate u with respect to v that is find
.
We have ![]()
By substituting x = tan θ, we have


[∵ sec2θ – tan2θ = 1]
![]()

![]()

But, cos2θ = 1 – 2sin2θ and sin2θ = 2sinθcosθ.


![]()
Given –1 < x < 1 ⇒ x ϵ (–1, 1)
However, x = tan θ
⇒ tan θ ϵ (–1, 1)
![]()
![]()
Hence, ![]()
![]()
On differentiating u with respect to x, we get
![]()
![]()
We know![]()
![]()
![]()
Now, we have ![]()
By substituting x = tan θ, we have
![]()
![]()
[∵ sec2θ – tan2θ = 1]

![]()
⇒ v = sin–1(2sinθcosθ)
But, sin2θ = 2sinθcosθ
⇒ v = sin–1(sin2θ)
However, ![]()
Hence, v = sin–1(sin2θ) = 2θ
⇒ v = 2tan–1x
On differentiating v with respect to x, we get
![]()
![]()
We know![]()
![]()
![]()
We have 

![]()
![]()
Thus, ![]()
Couldn't generate an explanation.
Generated by AI. May contain inaccuracies — always verify with your textbook.
(CBSE 2014)

(CBSE 2014)
(CBSE 2015)