Answer
$-4$ units per second
Work Step by Step
Given point $P(x,y)=P(x,2x) $ moves along the line $y=2x$
Let $ L$ be the given fixed-point $L (3,0) $
Let distance from the point $P (x,2x)$ to $L (3,0)$ be $z$.
The $z=\sqrt {(x-3) ^2+(2x-0)^2} $
$z=\sqrt {x^2-6x+9+4x^2} =\sqrt {5x^2-6x+9} $
Taking derivatives with respect to t
$\frac{dz}{dt}=\frac{d\sqrt {5x^2-6x+9}}{dt}$
Applying chain rule
$\frac{dz}{dt}=\frac{d\sqrt {5x^2-6x+9}}{dx} \frac{dx}{dt}$
$\frac{dz}{dx}= \frac{1}{2}[ {5x^2-6x+9}]^{-\frac{1}{2}} \frac{d(5x^2-6x+9)}{dx} \frac{dx}{dt}$
$ \frac{dz}{dx}=\frac{1}{2}\frac{10x-6}{\sqrt {5x^2-6x+9}} \frac{dz}{dx}$ ............. eq(1)
Putting $ x=3$ and $\frac{dx}{dt}=-2$ units/s in equation (1)
$ \frac{dz}{dx}=\frac{1}{2}\frac{10\times3-6} {\sqrt {5\times3^2-6\times3+9}} \times -2=-4$ units per second
