# Thread: Differentiation of Logarithmic and Exponential Functions

1. ## Differentiation of Logarithmic and Exponential Functions

Hello, I am having trouble differentiating this problem, it looks like the quotient rule is being used within the product rule, but I don't know how to combine the two of them.

f(x)= ln(x+1/x-1) I started using the product rule but then I got lost when I had to differentiate the second term, because this would be a quotient rule in this term. If someone could illustrate how to differentiate this, Thanks...

2. Originally Posted by kdogg121
Hello, I am having trouble differentiating this problem, it looks like the quotient rule is being used within the product rule, but I don't know how to combine the two of them.

f(x)= ln(x+1/x-1) I started using the product rule but then I got lost when I had to differentiate the second term, because this would be a quotient rule in this term. If someone could illustrate how to differentiate this, Thanks...
There are two ways:
The direct approach:
$f^{\prime}(x) = \frac{1}{\frac{x + 1}{x - 1}} \cdot \frac{(1)(x - 1) - (x + 1)(1)}{(x - 1)^2}$
etc.

or use the division property of logarithms:
$f(x) = ln(x + 1) - ln(x - 1)$
then take the derivative:
$f^{\prime}(x) = \frac{1}{x + 1} - \frac{1}{x - 1}$

The two methods have the same result. The second is obviously faster, but the first is better for algebra practice.

-Dan

3. I believe my professor wants us to use the first method you described, the answer in the book is: -2 / (x+1)(x-1) I don't see how this is obtained, I took the 1/(x+1)(x-1) times -2/(x-1)^2 so when I multiply the numerator matches the answer, however the denominator is different, I have ((x+1)/(x-1)) times (x-1)^2 and the book only has (x+1)(x-1) Are the (x-1) terms being cancelled when they are multiplied or what? Thanks again! Also what rules are you using? It doesnt look like the product rule, as there is no + sign between f g' and g f'.....So I am curious why in the first step you has 1/(x+1)/(x-1) isn't this term supposed to remain undifferentiated?

4. The derivative of log(x) is 1/x. He is using the chain rule.

5. Thank You for your help. I just would like to know if I have the right answer for the following problem:

f(x) = square root of ln t+t

I used the chain rule, got rid of the radical by raising the terms to the 1/2 power. I got as an answer: 1/2(ln t +t)^-1/2 ((1/t)+1) Is this correct?

The book has answer as: t+1 / 2t(square root of ln t +t) Thanks....

6. Originally Posted by kdogg121
Thank You for your help. I just would like to know if I have the right answer for the following problem:

f(x) = square root of ln t+t

I used the chain rule, got rid of the radical by raising the terms to the 1/2 power. I got as an answer: 1/2(ln t +t)^-1/2 ((1/t)+1) Is this correct? .... Yes

The book has answer as: t+1 / 2t(square root of ln t +t) Thanks....
$\frac12 (\ln(t)+t)^{-\frac12} \cdot \left(\frac1t + 1\right) = \frac{\frac1t + 1}{2 \cdot \sqrt{\ln(t)+t}} = \frac{\frac1t(1 + t)}{2 \cdot \sqrt{\ln(t)+t}} = \frac{t+1}{2t \cdot \sqrt{\ln(t)+t}}$