Derivatives of Exponential and Logarithm Functions — Question 7

Question 7 - Solution

We are given: $$f(x) = \ln\left( \frac{e^x + 1}{e^x - 1} \right)$$

Step 1: Use the derivative of the natural log of a quotient:

$$\frac{d}{dx} \ln\left( \frac{u(x)}{v(x)} \right) = \frac{d}{dx} \left[ \ln(u(x)) - \ln(v(x)) \right] = \frac{u'(x)}{u(x)} - \frac{v'(x)}{v(x)}$$

Here, $u(x) = e^x + 1$, and $v(x) = e^x - 1$

So,

$$f'(x) = \frac{d}{dx} \left[ \ln(e^x + 1) - \ln(e^x - 1) \right] = \frac{e^x}{e^x + 1} - \frac{e^x}{e^x - 1}$$

Step 2: Combine the two terms:

$$f'(x) = \frac{e^x}{e^x + 1} - \frac{e^x}{e^x - 1}$$

Get a common denominator:

$$f'(x) = e^x \left( \frac{1}{e^x + 1} - \frac{1}{e^x - 1} \right) = e^x \left( \frac{(e^x - 1) - (e^x + 1)}{(e^x + 1)(e^x - 1)} \right) = e^x \left( \frac{e^x - 1 - e^x - 1}{e^{2x} - 1} \right)$$

$$f'(x) = e^x \left( \frac{-2}{e^{2x} - 1} \right) = \boxed{ \frac{-2e^x}{e^{2x} - 1} }$$