The derivative of $\tan^{-1}\left(\dfrac{\sqrt{1+x^2} - 1}{x}\right)$ with respect to $\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)$ is...

If the derivative of $\tan^{-1}(a + bx)$ with respect to x at $x = 0$ is $1$, then $a^6 - b^3 = $

If $x^3 + y^3 = 6$ and $\dfrac{d^2y}{dx^2} \cdot \dfrac{d^2x}{dy^2} = \dfrac{m}{(xy)^n}$, where $m, n \in R$ then $\dfrac{m}{n} = $

If $x = \sin(t), y = \cos(pt)$, then $(1 - x^2)\dfrac{d^2 y}{dx^2} - x\dfrac{dy}{dx} = $