$\frac{d}{dx} \left( \lim_{y \to 2} \frac{1}{y-2} \left( \frac{1}{x} - \frac{1}{x+y-2} \right) \right) = $

Let $f : R \rightarrow R$ be defined as $f(x) = x^{3} + x - 5$. If $g(x)$ is a function such that $f(g(x)) = x, \forall x \in R$,then $g^{\prime}(63)$ is equal to

$f(x)$ and $g(x)$ are differentiable functions such that $\frac{f(x)}{g(x)} = c$,where $c$ is a non-zero constant. If $\frac{f^{\prime}(x)}{g^{\prime}(x)} = \alpha(x)$ and $\left(\frac{f(x)}{g(x)}\right)^{\prime} = \beta(x)$,then $\frac{\alpha(x) - \beta(x)}{\alpha(x) + \beta(x)} = $

If $y = \tan^{-1}\left( \frac{\sqrt{x} - x}{1 + x^{3/2}} \right)$,then $y'(1)$ is

Find the derivative: $\frac{d}{dx}(x e^{x^2}) = $

If $y = \frac{e^{2x} + e^{-2x}}{e^{2x} - e^{-2x}}$,then $\frac{dy}{dx} = $