Let $f(x) = x^m$,where $m$ is a non-negative integer. The value of $m$ such that the equality $f^{\prime}(a+b) = f^{\prime}(a) + f^{\prime}(b)$ is valid for all $a, b > 0$ is

If $f(1) = 1$ and $f'(1) = 3$,then the derivative of $f(f(f(x))) + (f(x))^2$ at $x = 1$ is:

Let the function satisfy the equation $f(x+y)=f(x)f(y)$ for all $x, y \in \mathbb{R}$,where $f(0) \neq 0$. If $f(5)=3$ and $f^{\prime}(0)=2$,then $f^{\prime}(5)$ is

People living on Mars,instead of the usual definition of derivative $D f(x)$,define a new kind of derivative,$D^*f(x)$ by the formula $D^*f(x) = \lim_{h \to 0} \frac{f^2(x + h) - f^2(x)}{h}$ where $f^2(x)$ means $[f(x)]^2$. If $f(x) = x \ln x$,then the value of $\left. D^*f(x) \right|_{x = e}$ is:

If $e^{x}=y+\sqrt{y^2-1}$,then $\frac{d y}{d x}=$

If $y = x^n \log x + x(\log x)^n$,then $\frac{dy}{dx} = $