Let $y(x) = (1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})$. Then $y'(x) - y''(x)$ at $x = -1$ is equal to:

$\mathop {\lim }\limits_{x \to 0} \frac{d}{{dx}}\left( {\frac{{{e^{{e^{{x^2}}}}} - e}}{x}} \right)$ is

Given the following properties of a function $f(x)$:
$(i)$ $f(x)$ is continuous and defined for all real numbers.
$(ii)$ $f'(-5) = 0$; $f'(2)$ is not defined and $f'(4) = 0$.
$(iii)$ $(-5, 12)$ is a point on the graph of $f(x)$.
$(iv)$ $f''(2)$ is undefined,but $f''(x)$ is negative everywhere else.
$(v)$ The signs of $f'(x)$ are given by the following number line:
$f'(x)$ is positive for $x < -5$,negative for $-5 < x < 2$,positive for $2 < x < 4$,and negative for $x > 4$.
On the possible graph of $y = f(x)$,we have:

If $f(x) = \begin{cases} x^{3}-3x+2, & x < 2 \\ x^{3}-6x^{2}+9x+2, & x \geq 2 \end{cases}$,then:

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\int_2^{\sec ^2 x} f(t) d t}{x^2-\frac{\pi^2}{16}}$ equals

Let $f(x)$ be a differentiable function,$f^{\prime}(x) > f(x)$ and $f(0) = 0$. Then