Question

# The derivative of $$tan^{-1} \left( \dfrac{\sqrt{1 + x^2} - 1}{x} \right )$$ with respect to $$tan^{-1} \left( \dfrac{2x \sqrt{1 - x^2}}{1 - 2x^2} \right )$$ at x = 0, is

A
18
B
14
C
12
D
1

Solution

## The correct option is C $$\dfrac{1}{4}$$Let $$t=\tan ^{ -1 }{ \left( \dfrac { \sqrt { 1+{ x }^{ 2 } } -1 }{ x } \right) }$$ Put $$x=\tan\theta$$ $$\Rightarrow t=\tan ^{ -1 }{ \left( \dfrac { \sqrt { 1+\tan ^{ 2 }{ \theta } } -1 }{ \tan { \theta } } \right) } \\ \Rightarrow t=\tan ^{ -1 }{ \left( \dfrac { \sec { \theta } -1 }{ \tan { \theta } } \right) } \\ \Rightarrow t=\tan ^{ -1 }{ \left( \dfrac { 1-\cos { \theta } }{ sin{ \theta } } \right) }$$ $$\Rightarrow t=\tan ^{ -1 }{ \left( \dfrac { 1-\dfrac { 1-\tan ^{ 2 }{ \dfrac { \theta }{ 2 } } }{ 1+\tan ^{ 2 }{ \dfrac { \theta }{ 2 } } } }{ \dfrac { 2\tan { \dfrac { \theta }{ 2 } } }{ 1+\tan ^{ 2 }{ \dfrac { \theta }{ 2 } } } } \right) }$$ $$\Rightarrow t=\tan ^{ -1 }{ \left( \tan { \dfrac { \theta }{ 2 } } \right) } =\dfrac { \theta }{ 2 } \\ \Rightarrow t=\dfrac { \tan ^{ -1 }{ x } }{ 2 }$$    .....(i) $$u=\tan ^{ -1 }{ \left( \dfrac { 2x\sqrt { 1-{ x }^{ 2 } } }{ 1-2{ x }^{ 2 } } \right) }$$ Put $$x=\sin\theta$$ $$u=\tan ^{ -1 }{ \left( \dfrac { 2\sin { \theta } \sqrt { 1-\sin ^{ 2 }{ \theta } } }{ 1-2\sin ^{ 2 }{ \theta } } \right) } \\ u=\tan ^{ -1 }{ \left( \dfrac { 2\sin { \theta } \cos { \theta } }{ \cos { 2\theta } } \right) } \\ u=\tan ^{ -1 }{ \left( \tan { 2\theta } \right) } \\ u=2\theta \\ \Rightarrow u=2\tan ^{ -1 }{ x } \\ \dfrac { du }{ dx } =\dfrac { 2 }{ \left( 1+{ x }^{ 2 } \right) }$$    .......(ii) Using $$(i)$$ and $$(ii)$$ $$\dfrac { dt }{ du } =\dfrac { \dfrac { 1 }{ 2\left( 1+{ x }^{ 2 } \right) } }{ \dfrac { 2 }{ \left( 1+{ x }^{ 2 } \right) } } =\dfrac { 1 }{ 4 }$$ So option B is correct.Mathematics

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