If \( f(x)=\frac{2 x^{2} \tan x}{\sec x} \), find \( f^{\prime}(x)=4 x \tan (x) \cos (x)+2 x^{2} \sec ^{2}(x) \cos (x)-2 x^{2} \frac{\sin ^{2}(x)}{\cos (x)} \) Find \( f^{\prime}(4)= \)

To find \( f'(4) \) for the function \( f(x) = \frac{2x^2 \tan x}{\sec x} \), we first need to evaluate the derivative \( f'(x) \) at \( x = 4 \). Given the expression for the derivative: \[ f'(x) = 4x \tan(x) \cos(x) + 2x^2 \sec^2(x) \cos(x) - 2x^2 \frac{\sin^2(x)}{\cos(x)} \] we will substitute \( x = 4 \) into this expression. 1. **Calculate \( \tan(4) \)**, \( \sec(4) \), and \( \cos(4) \): - \( \tan(4) = \tan(4) \) (we will compute this value numerically later) - \( \sec(4) = \frac{1}{\cos(4)} \) - \( \cos(4) = \cos(4) \) (we will compute this value numerically later) 2. **Substituting \( x = 4 \) into \( f'(x) \)**: \[ f'(4) = 4(4) \tan(4) \cos(4) + 2(4^2) \sec^2(4) \cos(4) - 2(4^2) \frac{\sin^2(4)}{\cos(4)} \] Simplifying the terms: - \( 4(4) = 16 \) - \( 2(4^2) = 32 \) Thus, we have: \[ f'(4) = 16 \tan(4) \cos(4) + 32 \sec^2(4) \cos(4) - 32 \frac{\sin^2(4)}{\cos(4)} \] 3. **Using trigonometric identities**: - \( \sec^2(4) = 1 + \tan^2(4) \) Therefore: \[ f'(4) = 16 \tan(4) \cos(4) + 32 (1 + \tan^2(4)) \cos(4) - 32 \frac{\sin^2(4)}{\cos(4)} \] 4. **Calculating \( \sin(4) \) and \( \cos(4) \)**: Using a calculator or numerical methods: - \( \sin(4) \approx -0.7568 \) - \( \cos(4) \approx -0.6536 \) - \( \tan(4) = \frac{\sin(4)}{\cos(4)} \approx \frac{-0.7568}{-0.6536} \approx 1.1578 \) 5. **Calculating \( \sec^2(4) \)**: \[ \sec^2(4) = \frac{1}{\cos^2(4)} \approx \frac{1}{(-0.6536)^2} \approx 2.3398 \] 6. **Substituting these values back into \( f'(4) \)**: \[ f'(4) \approx 16(1.1578)(-0.6536) + 32(2.3398)(-0.6536) - 32 \frac{(-0.7568)^2}{-0.6536} \] Calculating each term: - First term: \( 16(1.1578)(-0.6536) \approx -16.1576 \) - Second term: \( 32(2.3398)(-0.6536) \approx -48.7464 \) - Third term: \( 32 \frac{0.5712}{-0.6536} \approx -27.872 \) Combining these: \[ f'(4) \approx -16.1576 - 48.7464 + 27.872 \approx -37.032 \] Thus, the final answer is: \[ \boxed{-37.032} \]