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Given $f(x) = |\tan 2x|$. We need to find $f'(x)$ at $x = \frac{\pi}{3}$. First, we need to determine if $\tan 2x$ is positive or negative at $x = \frac{\pi}{3}$.
At $x = \frac{\pi}{3}$, $2x = \frac{2\pi}{3}$. Since $\frac{2\pi}{3}$ is in the second quadrant, $\tan \frac{2\pi}{3}$ is negative. Therefore, $\tan 2x < 0$ at $x = \frac{\pi}{3}$.
Since $\tan 2x < 0$ at $x = \frac{\pi}{3}$, we can rewrite $f(x)$ as $f(x) = -\tan 2x$ in a neighborhood around $x = \frac{\pi}{3}$.
Now, we differentiate $f(x) = -\tan 2x$ with respect to $x$: $$f'(x) = -\frac{d}{dx}(\tan 2x) = -(\sec^2 2x) \cdot 2 = -2\sec^2 2x$$
Now, we evaluate $f'(x)$ at $x = \frac{\pi}{3}$: $$f'\left(\frac{\pi}{3}\right) = -2\sec^2\left(2 \cdot \frac{\pi}{3}\right) = -2\sec^2\left(\frac{2\pi}{3}\right)$$ Since $\sec\left(\frac{2\pi}{3}\right) = \frac{1}{\cos\left(\frac{2\pi}{3}\right)} = \frac{1}{-\frac{1}{2}} = -2$, we have: $$f'\left(\frac{\pi}{3}\right) = -2(-2)^2 = -2(4) = -8$$
Final Answer: -8
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