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

For the function $f(x) = x^2 - 6x + 8$ where $2 \le x \le 4$,the value of $x$ for which $f'(x)$ vanishes is:

For the function $f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1$,prove that $f^{\prime}(1) = 100 f^{\prime}(0)$.

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 $y = \sqrt{\sin \sqrt{x}}$,then $\frac{dy}{dx} = $

Let $f$ and $g$ be differentiable functions on $R$ such that $f \circ g$ is the identity function. If for some $a, b \in R$,$g^{\prime}(a) = 5$ and $g(a) = b$,then $f^{\prime}(b)$ is equal to