Derivatives of Hyperbolic Functions — Question 3

Question 3 - Solution

We are given: $$f(x) = \tanh(x)\sinh(x)$$

This is a product of two functions, so we apply the product rule: $$f'(x) = \tanh'(x)\sinh(x) + \tanh(x)\sinh'(x)$$

Recall the derivatives: $$\tanh'(x) = \text{sech}^2(x), \quad \sinh'(x) = \cosh(x)$$

Substitute: $$f'(x) = \text{sech}^2(x)\sinh(x) + \tanh(x)\cosh(x)$$

Now simplify. Recall: $$\tanh(x) = \frac{\sinh(x)}{\cosh(x)} \Rightarrow \tanh(x)\cosh(x) = \sinh(x)$$

So: $$f'(x) = \text{sech}^2(x)\sinh(x) + \sinh(x) = \sinh(x)\left( \text{sech}^2(x) + 1 \right)$$

Final Answer: $$\boxed{f'(x) = \sinh(x)\left( \text{sech}^2(x) + 1 \right)}$$