http://i.imgur.com/OgOEINK.jpg
You can see the question and my answer/method to the question on the image.
However, the section i circled in red is what i am confused about. Would the ln (x-1)^2; which is placed there because we used the quotient rule for the question, cancel out both the ln (x-1) and ln (x+2) in the numerator?
The only reason why i did that was because it is the only way i was lead to the correct answerl
-3 / [(x+2) (x-1)]
Thanks.
You can see the question and my answer/method to the question on the image.
However, the section i circled in red is what i am confused about. Would the ln (x-1)^2; which is placed there because we used the quotient rule for the question, cancel out both the ln (x-1) and ln (x+2) in the numerator?
The only reason why i did that was because it is the only way i was lead to the correct answerl
-3 / [(x+2) (x-1)]
Thanks.
-
ln(a / b) does not equal ln(a) / ln(b), from the property of logs, split the log into two separate logs through subtraction:
y = ln(x + 2) - ln(x - 1)
Now take the derivative of each term:
dy/dx = [1/(x + 2) * 1] - [1/(x - 1) * 1]
dy/dx = 1 / (x + 2) - 1 / (x - 1)
Optionally, you can make the denominators common and combine them into a single fraction:
dy/dx = -3 / (x - 1)(x + 2)
y = ln(x + 2) - ln(x - 1)
Now take the derivative of each term:
dy/dx = [1/(x + 2) * 1] - [1/(x - 1) * 1]
dy/dx = 1 / (x + 2) - 1 / (x - 1)
Optionally, you can make the denominators common and combine them into a single fraction:
dy/dx = -3 / (x - 1)(x + 2)