Given that y=1+ln(2x-3), obtain an expression for dy/dx and...?

Question

Given that y=1+ln(2x-3), obtain an expression for dy/dx and hence find, in terms of p, the approximate value of y when x=2+p, where p is small

Answer 1

So first for dy/dx, you just have to apply the chain rule to the natural log portion since the derivative of 1 is 0. To apply the chain rule, you need to take the derivative of the outside function (ln) leaving the inside (2x-3) the same, then multiply by the derivative of the inside (the derivative of 2x-3 is 2). So we end up with this:

dy/dx = (1/(2x-3)) * 2

Or simplifying gives

dy/dx = 2/(2x-3)

To approximate y when x = 2+p for p small, this is really just asking you to calculate the Taylor polynomial of this function around 2. Since p is small, we only need the Taylor polynomial of degree 1. So viewing y as a function of x, say y(x), recall that the Taylor polynomial formula as a function of p about 2 is

F(p) = y(2) + dy/dx(2)(p-2)

Where y(2) means 'plug in 2 for x in y = 1 + ln(2x-3)' and dy/dx(2) means 'plug in 2 for x in dy/dx) = ...'

So just using the above formulas we have:

y(2) = 1 + ln(2(2) - 3) = 1 + ln(4-3) = 1 + ln(1) = 1

since ln(1) = 0. And also

dy/dx(2) = 2/(2(2) - 3) = 2 / (4-3) = 2

So we have

F(p) = 1 + 2(p-2)

I hope this helps.

Answer 2

dy/dx=du/u=2/(2x-3)

dunno about the rest.