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
= 1/(x-1)
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
To find the derivative of a function involving natural logarithm (ln), we can use the chain rule.
a) Let's find the derivative of y = ln(x - 1).
Step 1: Identify the inner function.
In this case, the inner function is x - 1.
Step 2: Calculate the derivative of the inner function.
The derivative of x - 1 is simply 1.
Step 3: Apply the chain rule.
The derivative of ln(u) with respect to u is 1/u, where u represents the inner function.
So, dy/dx = (1)/(x - 1) or 1/(x - 1).
It seems there was a mistake in your teacher's answer for part a). The correct derivative is 1/(x - 1), not (2x)/(x^2 - 1).
b) Now let's find the derivative of y = 3 ln x - ln (1/x).
Step 1: Simplify the expression.
Since ln(1/x) is the same as ln(1) - ln(x), we can rewrite the expression as:
y = 3 ln x - ln 1 + ln x = 3 ln x + ln x = 4 ln x.
Step 2: Apply the derivative of ln x.
The derivative of ln x is 1/x.
Step 3: Calculate the derivative.
dy/dx = 4 * (1/x) = 4/x.
According to the correct calculations, your teacher's answer for part b) is correct. The derivative of y = 3 ln x - ln (1/x) is 4/x.