If $y = \tan^{-1} \sqrt{\frac{1 - \cos x}{1 + \cos x}}$,then the values of $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ respectively are

$f(x) = e^x \sin x$,then $f^{(6)}(x)$ is equal to :

Let $f : S \rightarrow S$ where $S =(0, \infty)$ be a twice differentiable function such that $f(x+1) = xf(x)$. If $g : S \rightarrow R$ is defined as $g(x) = \log_{e} f(x)$,then the value of $|g''(5) - g''(1)|$ is equal to:

If $f$ is twice differentiable such that $f''(x) = -f(x)$,$f'(x) = g(x)$,$h'(x) = [f(x)]^2 + [g(x)]^2$,and $h(0) = 2$,$h(1) = 4$,then the equation $y = h(x)$ represents:

If $y^{1/4} + y^{-1/4} = 2x$,and $(x^2 - 1) \frac{d^2y}{dx^2} + \alpha x \frac{dy}{dx} + \beta y = 0$,then $|\alpha - \beta|$ is equal to ...... .

If $ y = (\tan^{-1} x)^2 $,then $ (x^2 + 1)^2 y_2 + 2x(x^2 + 1) y_1 $ is equal to