If $f:R \to R$ is a differentiable function and $f(2) = 6$,then $\lim_{x \to 2} \int_{6}^{f(x)} \frac{2t \, dt}{x - 2}$ is

Differentiate the function with respect to $x$: $\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$,where $0 < x < \frac{\pi}{2}$.

Let $f : R \rightarrow R$ be a differentiable function and $f(1) = 4$. Then the value of $\lim_{x \rightarrow 1} \int_{4}^{f(x)} \frac{2t \, dt}{x - 1}$ is:

If $y = \sin^{-1}\left(\frac{3x}{2} - \frac{x^3}{2}\right)$,then $\frac{dy}{dx}$ is equal to

Derivative of $\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ with respect to $\tan ^{-1} x$ for $-1 < x < 1$ is:

If $f(x)=e^x$,$g(x)=\sin^{-1} x$ and $h(x)=f(g(x))$,then $\frac{h^{\prime}(x)}{h(x)}$ is