what is the derivative of ln(x-1)?
If you are studying the calculus of logarithms your question surprises me, since that is one of the most basic ones in that topic.
in general if y = ln("anything")
then y' = 1/"anything" * derivative of "anything"
just apply this rule and let me know what you got.
To find the derivative of ln(x-1), we can use the chain rule. The chain rule states that if we have a composite function, such as ln(g(x)), the derivative can be found by taking the derivative of the outer function multiplied by the derivative of the inner function.
In this case, the outer function is ln(u) and the inner function is g(x) = x-1.
First, let's find the derivative of the inner function:
g'(x) = 1
Now, let's find the derivative of the outer function:
ln'(u) = 1/u
Finally, we can apply the chain rule and multiply the derivatives together:
(ln(g(x)))' = (1/u) * (g'(x))
= (1/(x-1)) * 1
= 1/(x-1)
Therefore, the derivative of ln(x-1) is 1/(x-1).