I can't see any errors on the steps. But Step 3 makes me doubt my answer.
1 Answer
$\begingroup$The product rule: $\frac{d}{dx}[f(x) \cdot g(x)] = f'(x)g(x) +f(x)g'(x)$.
It's technically not actually always true. For example, try $f(x) = |x| = g(x)$ at $x = 0$.
The catch is that $f'(x)$ and $g'(x)$ must be defined. If they are, then it is true.
So the assumption in the above problem occurs at step 3. It assumes $\frac{d}{dx}[\sec(x)]$ is defined.
Edit: It seems there is debate of whether the fact "$\frac{d}{dx}[\sec(x)]$ is defined when sec$(x)$ is defined" is actually an assumption for this question. Consider if the question had been this one instead:
Below is an attempt to derive the derivative of $ f(x) = \begin{cases} 1, & \text{if $x \in \mathbb{Q}$} \\ -1, & \text{if $x \notin \mathbb{Q}$} \end{cases}\Bigr\}$ using the product rule. In which step, if any, does an error first appear?
Step 1) $f(x) \cdot f(x) = 1$
Step 2) $\frac{d}{dx}[f(x) \cdot f(x)] = 0$
Step 3) $f'(x)f(x) + f(x)f'(x) = 0$
Step 4) $f'(x) = 0$
$\endgroup$ 8
