Question

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

Answer 1

`y=(\ln x)/(3x-6)`

Differentiate both sides with respect to
`x`:


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

When
`x=1`, you have


`(\mathrm dy)/(\mathrm dx)=\frac{\frac{3-6}1-3\ln1}{(3-6)^2}=\frac{-3}9=-\frac13`

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


`(\mathrm dy)/(\mathrm dt)=\frac{\frac{3x-6}x-3\ln x}{(3x-6)^2}(\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
`(\mathrm dy)/(\mathrm dt)=-0.1`. So when
`x=1`, you have


`-0.1=\frac{\frac{3-6}1-3\ln1}{(3-6)^2}(\mathrm dx)/(\mathrm dt)`

`-0.1=-\frac13(\mathrm dx)/(\mathrm dt)`

`(\mathrm dx)/(\mathrm dt)=0.3`

where the unit is again units/second.