If $$y = {A \over x} + B{x^2}$$, then $${x^2}{{{d^2}y} \over {d{x^2}}}$$ =

Let $$f(x) = ta{n^{ - 1}}x$$. Then $$f'(x) + f''(x)=0$$, when x is equal to

If $$y = {\tan ^{ - 1}}{{\sqrt {1 + {x^2}} - 1} \over x}$$, then y'(1) =

If $$y = 2{x^3} - 2{x^2} + 3x - 5$$, then for x = 2 and $$\Delta$$x = 0.1 the value of $$\Delta$$y is