Continuity and ...

$$\frac { d }{ dx } (\sin ^{ -1 }{ \{ \frac { \sqrt { 1+x } +\sqrt { 1-x }  }{ 2 } \}  } )=$$

$$dx\left\{ sin{  }^{ -1 }(\frac { 5x+12\sqrt { 1-x{  }^{ 2 } }  }{ 13 } ) \right\} =$$

If $$y=\cos ^{ -1 }{ (\frac { x-{ x }^{ -1 } }{ x+{ x }^{ -1 } } ) } \quad then\quad \frac { dy }{ dx } =$$

The number of point at which the function F(x)= max$$\left\{ a-x,a+x,b \right\} -\infty <x<\infty ,0<a<b$$ cannot be differentiable is 

If $$x=\theta-\frac{1}{\theta},y=\theta+\frac{1}{\theta}$$ then $$\dfrac{dy}{dx}$$=

Let fbe twice differentiable function such that $${ g }^{ 1 }\left( x \right) =-f\left( x \right) and\quad \quad \quad \quad \quad { f }^{ 1 }\left( x \right) =g\left( x \right) ,\\ h(x)={ \left( f\left( x \right)  \right)  }^{ 2 }+{ \left( g\left( x \right)  \right)  }^{ 2 }.\quad Ifh(5)=11,\quad thenh(10)\quad is$$

If $$y=\tan ^{ -1 }{ [x+\sqrt { 1+{ x }^{ 2 } } ] } $$ then $$\frac { dy }{ dx } =$$

$$\frac { d[\sec ^{ -1 }{ (\sin { x } +{ x }^{ 2 }) } ] }{ dx } =$$

If $$y={ tan }^{ -1 }\left( \dfrac { 1-{ cos }^{ 2 }x }{ 1+{ cos }^{ 2 } }  \right) ,$$, then $$\dfrac { dy }{ xy } =$$

Solve $$\frac{d\tan ^{-1}}{dx} (\frac{5x + 1}{3 - x - 6x^2}) = $$