Derivatives of Hyperbolic Functions — Question 10

Question 10 - Solution

We are given: $$f(x) = \frac{\sinh(3x)}{\cosh(x)}$$

(a) Apply the quotient rule:

The quotient rule is: $$\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$$

Let: $$u = \sinh(3x), \quad v = \cosh(x)$$

Differentiate: $$u' = 3\cosh(3x), \quad v' = \sinh(x)$$

Apply the rule: $$f'(x) = \frac{3\cosh(3x)\cosh(x) - \sinh(3x)\sinh(x)}{\cosh^2(x)}$$

(b) Final simplified expression: $$\boxed{f'(x) = \frac{3\cosh(3x)\cosh(x) - \sinh(3x)\sinh(x)}{\cosh^2(x)}}$$