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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
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B
14
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C
12
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D
1
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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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