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

To find \( f'(4) \) for the function \( f(x) = \frac{2 x^2 \tan x}{\sec x} \), we first need to simplify the expression for \( f(x) \). Recall that \( \sec x = \frac{1}{\cos x} \), so we can rewrite \( f(x) \) as: \[ f(x) = 2 x^2 \tan x \cdot \cos x \] Now, we can differentiate \( f(x) \) using the product rule. Let \( u = 2 x^2 \) and \( v = \tan x \cos x \). Then: \[ f(x) = u v \] The derivative \( f'(x) \) using the product rule is: \[ f'(x) = u' v + u v' \] Calculating \( u' \): \[ u' = \frac{d}{dx}(2 x^2) = 4x \] Next, we need to find \( v' \). We have \( v = \tan x \cos x \). We can use the product rule again: \[ v' = \frac{d}{dx}(\tan x) \cos x + \tan x \frac{d}{dx}(\cos x) \] Calculating \( \frac{d}{dx}(\tan x) \) and \( \frac{d}{dx}(\cos x) \): \[ \frac{d}{dx}(\tan x) = \sec^2 x \] \[ \frac{d}{dx}(\cos x) = -\sin x \] Thus, \[ v' = \sec^2 x \cos x - \tan x \sin x \] Now substituting back into the expression for \( f'(x) \): \[ f'(x) = 4x \tan x \cos x + 2 x^2 \left( \sec^2 x \cos x - \tan x \sin x \right) \] Now we can simplify this expression: \[ f'(x) = 4x \tan x \cos x + 2 x^2 \sec^2 x \cos x - 2 x^2 \tan x \sin x \] Now we can evaluate \( f'(4) \). We need to calculate \( \tan(4) \), \( \sec(4) \), and \( \sin(4) \) using a calculator or tables: 1. Calculate \( \tan(4) \) 2. Calculate \( \sec(4) = \frac{1}{\cos(4)} \) 3. Calculate \( \sin(4) \) Using approximate values: - \( \tan(4) \approx 1.1578 \) - \( \cos(4) \approx -0.6536 \) so \( \sec(4) \approx -1.529 \) - \( \sin(4) \approx -0.7568 \) Now substituting these values into \( f'(4) \): \[ f'(4) = 4(4)(1.1578)(-0.6536) + 2(4^2)(-1.529)(-0.6536) - 2(4^2)(1.1578)(-0.7568) \] Calculating each term: 1. \( 4(4)(1.1578)(-0.6536) \approx -12.115 \) 2. \( 2(16)(-1.529)(-0.6536) \approx 20.116 \) 3. \( -2(16)(1.1578)(-0.7568) \approx 35.157 \) Now summing these results: \[ f'(4) \approx -12.115 + 20.116 + 35.157 \approx 43.158 \] Thus, the final answer is: \[ \boxed{43.158} \]