Derivatives of Hyperbolic Functions — Question 2

Question 2 - Solution

The quotient rule gives

$$f'(x)=\frac{3\cosh(3x)\cosh x-\sinh(3x)\sinh x}{\cosh^2x}.$$

Use the product identities

$$\cosh(3x)\cosh x=\tfrac12(\cosh4x+\cosh2x),$$

$$\sinh(3x)\sinh x=\tfrac12(\cosh4x-\cosh2x).$$

Thus

$$\boxed{f'(x)=\frac{\cosh4x+2\cosh2x}{\cosh^2x}}.$$

At zero the derivative is $3$, agreeing with the original quotient-rule expression.