Show solutionHide solution+ Question 10 Let f(x)=sinh(3x)cosh(x)f(x) = \frac{\sinh(3x)}{\cosh(x)}. Differentiate f(x)f(x) using the quotient rule and known derivatives of hyperbolic functions. Simplify the expression for f′(x)f'(x) as much as possible. Question 10 - Solution We are given: f(x)=sinh(3x)cosh(x)f(x) = \frac{\sinh(3x)}{\cosh(x)} (a) Apply the quotient rule: The quotient rule is: (uv)′=u′v−uv′v2\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} Let: u=sinh(3x),v=cosh(x)u = \sinh(3x), \quad v = \cosh(x) Differentiate: u′=3cosh(3x),v′=sinh(x)u' = 3\cosh(3x), \quad v' = \sinh(x) Apply the rule: f′(x)=3cosh(3x)cosh(x)−sinh(3x)sinh(x)cosh2(x)f'(x) = \frac{3\cosh(3x)\cosh(x) - \sinh(3x)\sinh(x)}{\cosh^2(x)} (b) Final simplified expression: f′(x)=3cosh(3x)cosh(x)−sinh(3x)sinh(x)cosh2(x)\boxed{f'(x) = \frac{3\cosh(3x)\cosh(x) - \sinh(3x)\sinh(x)}{\cosh^2(x)}}