What is the value of $\frac{1}{2}$?

If $y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots + \frac{x^n}{n!}$,then $\frac{dy}{dx} = $

The value of the integral $\int_{-1}^{+1} \frac{1}{t^3} \, dt$ is:

If $y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + .....\infty$,then $\frac{dy}{dx} = $

$\frac{d}{dx} \log(\log x) = $

$\int \frac{dx}{{(2ax - x^2)}^{1/2}} = a^n \sin^{-1} \left( \frac{x}{a} - 1 \right)$. In this formula,$n =$ . . . . . . .