Derivative of $\tan^{-1}(f(x))$ What is derivative of $$\tan^{-1}\left(\frac{{\sqrt{4+x}+\sqrt{4-x}}}{\sqrt{4+x}-\sqrt{4-x}}\right).$$  So I tried to write it as $\tan(\tan^{-1}(...))$ to get the $f(x)=\frac{\pi}{4}+\tan^{-1}\left(\sqrt{\frac{4+x}{4-x}}\right)$ but still it's not better. Thanks help appreciated

$$\tan ^{ -1 } \left( \frac { \sqrt { 4+x } +\sqrt { 4-x }  }{ \sqrt { 4+x } -\sqrt { 4-x }  }  \right) =\tan ^{ -1 } \left( \frac { { \left( \sqrt { 4+x } +\sqrt { 4-x }  \right)  }^{ 2 } }{ \left( \sqrt { 4+x } -\sqrt { 4-x }  \right) \left( \sqrt { 4+x } +\sqrt { 4-x }  \right)  }  \right) =\\ =\tan ^{ -1 }{ \left( \frac { 4+x-2\sqrt { 16-{ x }^{ 2 } } +4-x }{ 2x }  \right) = } \tan ^{ -1 }{ \left( \frac { 8-2\sqrt { 16-{ x }^{ 2 } }  }{ 2x }  \right)  } \\ \frac { d }{ dx } \left( \tan ^{ -1 }{ \left( \frac { 8-2\sqrt { 16-{ x }^{ 2 } }  }{ 2x }  \right)  }  \right) =\frac { 1 }{ 1+{ \left( \frac { 8-2\sqrt { 16-{ x }^{ 2 } }  }{ 2x }  \right)  }^{ 2 } } \frac { \left( \frac { 2x }{ \sqrt { 16-{ x }^{ 2 } }  }  \right) 2x-2\left( 8-2\sqrt { 16-{ x }^{ 2 } }  \right)  }{ 4{ x }^{ 2 } } =\\ =\frac { 4{ x }^{ 2 } }{ 4{ x }^{ 2 }+{ \left( 8-2\sqrt { 16-{ x }^{ 2 } }  \right)  }^{ 2 } } \frac { 4{ x }^{ 2 }-16\sqrt { 16-{ x }^{ 2 } } +4\left( 16-{ x }^{ 2 } \right)  }{ 4{ x }^{ 2 }\left( \sqrt { 16-{ x }^{ 2 } }  \right)  } =\frac { 4{ x }^{ 2 }-16\sqrt { 16-{ x }^{ 2 } } +4\left( 16-{ x }^{ 2 } \right)  }{ \left( \sqrt { 16-{ x }^{ 2 } }  \right) \left( 4{ x }^{ 2 }+{ \left( 8-2\sqrt { 16-{ x }^{ 2 } }  \right)  }^{ 2 } \right)  } $$
