Derivatives of Exponential and Logarithm Functions — Question 9

Question 9

Let $a > 0$ be a real constant and define the function $$f(x) = \ln\!\left(\frac{e^{ax} + 1}{e^{ax} - 1}\right).$$

• Find the derivative $f'(x)$ in terms of $a$.

• Simplify your answer as much as possible.

• Determine the domain of $f(x)$.

• Show that $f'(x)$ can be written in the form $$f'(x) = -\frac{2a}{e^{ax} - e^{-ax}}.$$

• Hence, or otherwise, determine whether $f(x)$ is increasing or decreasing on its domain.

Question 9 - Solution

We are given $$f(x) = \ln\!\left(\frac{e^{ax} + 1}{e^{ax} - 1}\right), \quad a > 0.$$

(a) Differentiation

Rewrite the function using logarithmic identities: $$f(x) = \ln(e^{ax} + 1) - \ln(e^{ax} - 1).$$

Differentiate term by term using the chain rule: $$f'(x) = \frac{a e^{ax}}{e^{ax} + 1} - \frac{a e^{ax}}{e^{ax} - 1}.$$

(b) Simplification

Factor out $a e^{ax}$: $$f'(x) = a e^{ax}\left(\frac{1}{e^{ax} + 1} - \frac{1}{e^{ax} - 1}\right).$$

Combine the fractions: $$= a e^{ax} \left( \frac{(e^{ax} - 1) - (e^{ax} + 1)}{(e^{ax} + 1)(e^{ax} - 1)} \right) = a e^{ax} \left( \frac{-2}{e^{2ax} - 1} \right).$$

Thus, $$f'(x) = -\frac{2a e^{ax}}{e^{2ax} - 1}.$$

(c) Domain of $f(x)$

The logarithm requires: $$\frac{e^{ax} + 1}{e^{ax} - 1} > 0, \quad e^{ax} - 1 \neq 0.$$

Since $e^{ax} > 0$ for all $x$, the expression is positive only when $$e^{ax} - 1 > 0 \quad \Rightarrow \quad x > 0.$$

Hence, the domain is: $$\boxed{(0, \infty)}.$$

(d) Alternative Form

Multiply numerator and denominator of the derivative by $e^{-ax}$: $$f'(x) = -\frac{2a e^{ax}}{e^{2ax} - 1} = -\frac{2a}{e^{ax} - e^{-ax}}.$$

(e) Monotonicity

For $x > 0$, $$e^{ax} - e^{-ax} > 0, \quad a > 0.$$

Therefore, $$f'(x) < 0 \quad \text{for all } x > 0.$$

$$\boxed{\text{The function } f(x) \text{ is strictly decreasing on its domain.}}$$