We are given the function $f(x) = \cosh(6x - 7)$ and asked to find $f'(0)$.

1. Problem Description

We are given the function $f(x) = \cosh(6x - 7)$ and asked to find $f'(0)$.

2. Solution Steps

First, we need to find the derivative of $f(x)$ with respect to $x$. Recall that the derivative of $\cosh(x)$ is $\sinh(x)$. Also, remember the chain rule: if $f(x) = g(h(x))$, then $f'(x) = g'(h(x)) \cdot h'(x)$.

Applying the chain rule:

$f'(x) = \sinh(6x - 7) \cdot \frac{d}{dx}(6x - 7) = \sinh(6x - 7) \cdot 6 = 6\sinh(6x - 7)$.

Next, we need to find $f'(0)$.

$f'(0) = 6\sinh(6(0) - 7) = 6\sinh(-7)$.

Recall that $\sinh(x) = \frac{e^x - e^{-x}}{2}$.

So, $\sinh(-7) = \frac{e^{-7} - e^7}{2}$.

$f'(0) = 6\sinh(-7) = 6 \cdot \frac{e^{-7} - e^7}{2} = 3(e^{-7} - e^7) \approx 3(0.00091188 - 1096.633) = 3(-1096.632) \approx -3289.896$.

3. Final Answer

The final answer is B. -3289.897