Continuity and ...

Let $$f$$ be a differentiable function for all $$x$$. If $$f(1) = -2$$ and $$f'(x) \ge 2$$ for $$x \in [1, 6]$$ then

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

If y = tan $$^{-1}$$ $$\dfrac{\sqrt{1 + x^2} -1}{x}$$, then$$\dfrac{dy}{dx}$$=

The function $$f\left( x \right) =\left( { x }^{ 2 }-1 \right) \left| { x }^{ 2 }-3x+2 \right| +\cos { \left( \left| x \right|  \right)  }$$ is not differentiable at

If $$y = \cos^{-1}\left(\dfrac{x - x^{-1}}{x + x^{-1}}\right)$$, then $$\dfrac{dy}{dx}$$ is equal to

if $$f(x)=a\left| sinx \right| +{ be }^{ x }+c\left| { x }^{ 3 } \right| $$, where $$ a,b,c \epsilon$$ R, is differentiable at $$x=0$$ then

If $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$$, then $$\dfrac{d^{2}y}{dx^{2}}=$$

$$f(x)=min {1,cos\,x,1-sin\,x}, -\pi \leq x \leq \pi$$ then

If $$Cos^{-1}\left ( \frac{x^2-y^2}{x^2+y^2} \right )=a$$, then $$\frac{dy}{dx}=$$

Let  $$f : R \rightarrow R$$  be a twice differentiable function satisfying  $$f^{ { \prime \prime  } }(x)=5f^{ { \prime  } }(x)+6f\geq 0,\forall x\geq 0,f(0)=1$$  and  $$f ^ { \prime } ( 0 ) = 0 .$$  if  $$f ( x )$$  satisfies  $$f(x)\geq a.h(bx)-b\cdot \overline { h } (ax),\forall x\geq 0,$$  then  $$a + b$$  is equal to