Let $f(x) = \tan^{-1} x$. Then $f'(x) + f''(x) = 0$ when $x$ is equal to:

Let $f$ and $g$ be twice differentiable functions such that $f(x) \cdot g(x) = 1$ for all $x \in R$ and $f'$ and $g'$ are never zero. Then $\frac{f''(x)}{f(x)} + \frac{g''(x)}{g(x)}$ equals:

If $f(x)$ is a function such that $f^{\prime \prime}(x)+f(x)=0$ and $g(x)=[f(x)]^{2}+[f^{\prime}(x)]^{2}$ and $g(3)=8$,then $g(8)$ is equal to

If $x=\frac{1-\sqrt{y}}{1+\sqrt{y}}$, then $(x+1) \frac{d^2 y}{d x^2}+\left(\frac{3 \sqrt{y}+1}{\sqrt{y}}\right) \frac{d y}{d x}$ equals

If $y = \cos^2\left(\frac{5x}{2}\right) - \sin^2\left(\frac{5x}{2}\right)$,then $\frac{d^2y}{dx^2} =$

If $y = \operatorname{Sin}^{-1}\left(\frac{2x}{1+x^2}\right)$ and $\left(\frac{d^2y}{dx^2}\right)_{x=2} = k$,then $25k =$