Question

If \( f(x) = |\tan 2x| \), then find the value of \( f'(x) \) at \( x = \frac{\pi}{3} \).

Hint:

When differentiating functions involving absolute values, handle the positive and negative cases separately to account for the sign change in the derivative.

Answer 1

Step 1: Differentiate \( f(x) = |\tan 2x| \).
The absolute value function \( |u| \) is defined as: \[ |u| = \begin{cases} u, & \text{if } u>0,
-u, & \text{if } u<0. \end{cases} \] For the function \( f(x) = |\tan 2x| \), we differentiate separately depending on whether \( \tan 2x \) is positive or negative: \[ f'(x) = \begin{cases} \frac{d}{dx} (\tan 2x), & \text{if } \tan 2x>0,
\frac{d}{dx} (-\tan 2x), & \text{if } \tan 2x<0. \end{cases} \] The derivative simplifies to: \[ f'(x) = \begin{cases} 2 \sec^2 2x, & \text{if } \tan 2x>0,
-2 \sec^2 2x, & \text{if } \tan 2x<0. \end{cases} \] Step 2: Evaluate \( \tan 2x \) at \( x = \frac{\pi}{3} \).
Substitute \( x = \frac{\pi}{3} \): \[ \tan 2x = \tan \left(2 \cdot \frac{\pi}{3}\right) = \tan \frac{2\pi}{3}. \] Since \( \tan \frac{2\pi}{3} = -\sqrt{3} \), we have \( \tan 2x<0 \) at \( x = \frac{\pi}{3} \). Step 3: Substitute into \( f'(x) \).
Since \( \tan 2x<0 \), the derivative is: \[ f'(x) = -2 \sec^2 2x. \] Now, calculate \( \sec^2 2x \) at \( x = \frac{\pi}{3} \): \[ \sec 2x = \sec \frac{2\pi}{3} = -\sec \frac{\pi}{3} = -2 \quad \Rightarrow \quad \sec^2 2x = (-2)^2 = 4. \] Thus: \[ f'(x) = -2 \cdot 4 = -8. \] Step 4: Final Answer.
The value of \( f'(x) \) at \( x = \frac{\pi}{3} \) is: \[ \boxed{-8}. \]