Derivatives of Hyperbolic Functions — Question 8

Question 8 - Solution

We are given: $$f(x) = \frac{\sinh(3x)}{x^2 + 1}$$

(a) Find $f'(x)$:

This is a quotient, so we use the quotient rule: $$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$$

Let: $$g(x) = \sinh(3x), \quad h(x) = x^2 + 1$$

Then: $$g'(x) = 3\cosh(3x), \quad h'(x) = 2x$$

So: $$f'(x) = \frac{3\cosh(3x)(x^2 + 1) - \sinh(3x)(2x)}{(x^2 + 1)^2}$$

$$\boxed{ f'(x) = \frac{3(x^2 + 1)\cosh(3x) - 2x\sinh(3x)}{(x^2 + 1)^2} }$$

(b) Evaluate at $x = 0$:

First note: $$\cosh(0) = 1, \quad \sinh(0) = 0$$

Then: $$f'(0) = \frac{3(0^2 + 1)\cdot 1 - 0}{(0^2 + 1)^2} = \frac{3}{1} = 3$$

$$\boxed{f'(0) = 3}$$