y=ln(x+1) +ln(x-1) find dy/dx

I have trouble with this problem since i'm not that familar with finding the derivative when they involve ln. Please show me how to solve this problem. I should have gotten 2x/(x^2-1)

The derivative of ln(x) is 1/x

So, you obtain

dy/dx = 1/(x+1) + 1/(x-1)

Adding the fractions gives you the desired answer 2x/(x^2-1) .

To find the derivative of y = ln(x+1) + ln(x-1), you can use the properties of logarithmic differentiation.

Step 1: Identify the different logarithmic terms in the equation.
In this case, we have two logarithmic terms: ln(x+1) and ln(x-1).

Step 2: Use the rule that the derivative of ln(x) is 1/x.
The derivative of ln(x+1) would be 1/(x+1), and the derivative of ln(x-1) would be 1/(x-1).

Step 3: Apply the sum rule for derivatives.
Since we are adding the two logarithmic terms together, we can simply add their respective derivatives.

dy/dx = 1/(x+1) + 1/(x-1)

Step 4: Simplify the expression.
To simplify the expression, you can find a common denominator for the fractions:

dy/dx = (x-1)/(x-1)(x+1) + (x+1)/(x+1)(x-1)

Now you can combine the fractions by adding the numerators (since they have the same denominator):

dy/dx = (x-1 + x+1)/(x^2 - 1)

Simplifying further:

dy/dx = (2x)/(x^2 - 1)

Therefore, the derivative of y = ln(x+1) + ln(x-1) is dy/dx = 2x/(x^2 - 1).