If $x=\log_e t$ and $y=\dfrac{1}{t}$, then $\dfrac{d^2y}{dx^2}$ at $t=0.1$ is

$\frac{d^2}{dx^2} \left\{ \det \begin{bmatrix} x^3 & x \\ 2 & e^x \end{bmatrix} \right\}$ equals

If $y = \sin^{-1} x + \sin^{-1} \sqrt{1-x^2}, x \in (-1, 0)$, then $\frac{dy}{dx}$ is equal to

Let $e^y(x+1) = 1$. Then which of the following are TRUE?

(A) $\frac{d^2y}{dx^2} = -\frac{1}{(x+1)^2}$

(B) $\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2$

(C) $\left.\frac{d^2y}{dx^2}\right|_{x=0} = -1$

(D) $\left.\frac{d^2y}{dx^2}\right|_{x=0} = 1$

(E) $\left.\frac{d^2y}{dx^2}\right|_{x=1} = \frac{1}{4}$

Choose the correct answer from the options given below:

If $y = (log x)^{(log x)}$, $x > 1$ then $\frac{dy}{dx}$ is equal to

If $y = \log\left[\frac{x^2}{e^2}\right]$ then value of $\frac{d^2y}{dx^2}$ is :

$\frac{d}{dx}\left(e^{2\log_e x^3}\right)$ equals