If $y=(x-1)(x+2)\left(x^2+5\right)\left(x^4+8\right)$, then $\lim _{x \rightarrow-1}\left(\frac{d y}{d x}\right)$ is equal to

If $f(x)=\left\{\begin{array}{cc}a x^2+b x-\frac{13}{8}, & x \leq 1 \\ 3 x-3, & 1 < x \leq 2 \text { is differentiable } \\ b x^3+1, & x > 2\end{array}\right.$ $\forall x \in R$, then $a-b$ is equal to

$A$ is a point on the circle with radius 8 and centre at $O$. A particle $P$ is moving on the circumference of the circle starting from $A . M$ is the foot of the perpendicular from $P$ on $O A$ and $\angle P O M=\theta$. When $O M$ $=4$ and $\frac{d \theta}{d t}=6$ radians $/ \mathrm{sec}$, then the rate of change of $P M$ is (in units/sec)

If the length of the sub-tangent at any $P$ on a curve is proportional to the abscissa of the point $P$, then the equation of that curve is ( $C$ is an arbitrary constant)