a) y =ln(x-1) find dy/dx=?
b) y=3 ln x - ln (1/x) where x >0, dy/dx= ?
My teacher says the answers are a) (2x)/(x^2-1) and b) 4/x
I am not sure how you find the derivative when they involve ln. Could you please explain to me how to get these answers? I have a quiz on them soon so I just want to clearify how you would solve them. Thanks.
<<I am not sure how you find the derivative when they involve ln.>>
The derivative of ln x is 1/x. Use the "function of a function" rule when the log is of a function of x. For example:
y = ln (x-1)
Let u(x) = x-1
dy/dx = d (ln u)/du * du/dx = 1/u
You teacher is wrong on that one.
(2x)/(x^2-1)is the derivative of
ln (x^2 -1)
Perhaps you copied the problem wrong.
If y = 3 ln x - ln (1/x)
y = 3 ln x + ln x = 4 ln x
dy/dx = 4/x