If $2t = v^2$,then $\frac{dv}{dt}$ is equal to

Let $f$ be a differentiable function satisfying $f(x + 2y) = 2yf(x) + xf(y) - 3xy + 1$ for all $x, y \in R$ such that $f'(0) = 1$. Then $f(2)$ is equal to:

Consider the functions defined implicitly by the equation $y^3-3y+x=0$ on various intervals in the real line. If $x \in(-\infty,-2) \cup(2, \infty)$,the equation implicitly defines a unique real valued differentiable function $y=f(x)$. If $x \in(-2,2)$,the equation implicitly defines a unique real valued differentiable function $y=g(x)$ satisfying $g(0)=0$.
$1.$ If $f(-10 \sqrt{2})=2 \sqrt{2}$,then $f^{\prime \prime}(-10 \sqrt{2})=$
$(A)$ $\frac{4 \sqrt{2}}{7^3 3^2}$ $(B)$ $-\frac{4 \sqrt{2}}{7^3 3^2}$ $(C)$ $\frac{4 \sqrt{2}}{7^3 3}$ $(D)$ $-\frac{4 \sqrt{2}}{7^3 3}$
$2.$ The area of the region bounded by the curves $y=f(x)$,the $x$-axis,and the lines $x=a$ and $x=b$,where $-\infty < a < b < -2$,is
$(A)$ $\int_a^b \frac{x}{3\left((f(x))^2-1\right)} dx+bf(b)-af(a)$
$(B)$ $-\int_a^b \frac{x}{3\left((f(x))^2-1\right)} dx+bf(b)-af(a)$
$(C)$ $\int_a^b \frac{x}{3\left((f(x))^2-1\right)} dx-bf(b)+af(a)$
$(D)$ $-\int_a^b \frac{x}{3\left((f(x))^2-1\right)} dx-bf(b)+af(a)$
$3.$ $\int_{-1}^1 g^{\prime}(x) dx=$
$(A)$ $2g(-1)$ $(B)$ $0$ $(C)$ $-2g(1)$ $(D)$ $2g(1)$
Give the answer for questions $1, 2$ and $3.$

If $y = \sqrt{\sin^{-1} x + y}$,then $\frac{dy}{dx} = $ . . . . . . . (where,$x \in (0, 1)$)

If $\sqrt{\frac{y}{x}}+4 \sqrt{\frac{x}{y}}=4$, then $\frac{d y}{d x}=$

If $\sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}}=2$, then $\frac{dy}{dx}$ is equal to