Derivatives of Hyperbolic Functions — Question 4

Question 4

Let $f(x) = \frac{\cosh^2(x) - 1}{\sinh(x)}$.

• Simplify the expression before differentiating.

• Compute $f'(x)$.

• Identify any points where $f'(x)$ is undefined.

Question 4 - Solution

(a) Simplification:

We start with: $$f(x) = \frac{\cosh^2(x) - 1}{\sinh(x)}$$

Use the identity: $$\cosh^2(x) - 1 = \sinh^2(x)$$

Thus: $$f(x) = \frac{\sinh^2(x)}{\sinh(x)} = \sinh(x)$$

(b) Differentiate:

$$f'(x) = \frac{d}{dx} \sinh(x) = \cosh(x)$$

(c) Points of Non-Differentiability:

Since $f(x) = \sinh(x)$ is differentiable for all real $x$, and the simplification is valid as long as the denominator of the original function $\sinh(x) \neq 0$, we must consider the domain of the original function.

The original function is undefined at: $$\sinh(x) = 0 \Rightarrow x = 0$$

So although $f(x) = \sinh(x)$ is differentiable everywhere, the original form $f(x) = \frac{\cosh^2(x) - 1}{\sinh(x)}$ is undefined at $x = 0$, and thus $f$ is not defined or differentiable there in its original form.

Conclusion: $$\boxed{f'(x) = \cosh(x), \quad f'(x) \text{ undefined at } x = 0}$$