Question

How to do question 1?Had no idea how to do both parts

Answer 1

$y=\dfrac{\ln x}{3x-6}$

Differentiate both sides with respect to $x$:

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{3x-6}x-3\ln x}{(3x-6)^2}$

When $x=1$, you have

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{3-6}1-3\ln1}{(3-6)^2}=\dfrac{-3}9=-\dfrac13$

For part (b), we now assume that $x$ and $y$ are functions of an independent variable, which we'll call $t$ (for time). Now differentiating both sides with respect to $t$, we have

$\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{\frac{3x-6}x-3\ln x}{(3x-6)^2}\dfrac{\mathrm dx}{\mathrm dt}$

where the chain rule is used on the right side. We're told that $y$ is decreasing at a constant rate of 0.1 units/second, which translates to $\dfrac{\mathrm dy}{\mathrm dt}=-0.1$. So when $x=1$, you have

$-0.1=\dfrac{\frac{3-6}1-3\ln1}{(3-6)^2}\dfrac{\mathrm dx}{\mathrm dt}$
$-0.1=-\dfrac13\dfrac{\mathrm dx}{\mathrm dt}$
$\dfrac{\mathrm dx}{\mathrm dt}=0.3$

where the unit is again units/second.