If \( f(x)=\frac{\tan x-4}{\sec x} \) \( f^{\prime}(x)=\square \) \( f^{\prime}(5)=\square \)

To find the derivative \( f'(x) \) of the function \( f(x) = \frac{\tan x - 4}{\sec x} \), we can use the quotient rule. The quotient rule states that if \( f(x) = \frac{g(x)}{h(x)} \), then \[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \] In our case, let \( g(x) = \tan x - 4 \) and \( h(x) = \sec x \). First, we need to find \( g'(x) \) and \( h'(x) \): 1. The derivative of \( g(x) = \tan x - 4 \) is: \[ g'(x) = \sec^2 x \] 2. The derivative of \( h(x) = \sec x \) is: \[ h'(x) = \sec x \tan x \] Now we can apply the quotient rule: \[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \] Substituting \( g(x) \), \( g'(x) \), \( h(x) \), and \( h'(x) \): \[ f'(x) = \frac{\sec^2 x \cdot \sec x - (\tan x - 4) \cdot \sec x \tan x}{(\sec x)^2} \] This simplifies to: \[ f'(x) = \frac{\sec^3 x - (\tan x - 4) \sec x \tan x}{\sec^2 x} \] Now, we can simplify further: \[ f'(x) = \frac{\sec^3 x}{\sec^2 x} - \frac{(\tan x - 4) \sec x \tan x}{\sec^2 x} \] This simplifies to: \[ f'(x) = \sec x - \frac{(\tan x - 4) \tan x}{\sec x} \] Now, we can express \( \sec x \) in terms of \( \cos x \): \[ f'(x) = \sec x - (\tan^2 x - 4 \tan x) \cos x \] Next, we need to evaluate \( f'(5) \). First, we calculate \( \tan(5) \) and \( \sec(5) \): 1. Calculate \( \tan(5) \) and \( \sec(5) \): - \( \tan(5) \) is approximately \( -0.9589 \) - \( \sec(5) = \frac{1}{\cos(5)} \) is approximately \( 1.2837 \) Now substituting these values into \( f'(x) \): \[ f'(5) = \sec(5) - \left( \tan^2(5) - 4 \tan(5) \right) \cos(5) \] Calculating \( \tan^2(5) \): \[ \tan^2(5) \approx (-0.9589)^2 \approx 0.9195 \] Now substituting into the expression: \[ f'(5) = 1.2837 - (0.9195 - 4(-0.9589)) \cos(5) \] Calculating \( 4(-0.9589) \): \[ 4(-0.9589) \approx -3.8356 \] Now substituting back: \[ f'(5) = 1.2837 - (0.9195 + 3.8356) \cos(5) \] Calculating \( 0.9195 + 3.8356 \): \[ 0.9195 + 3.8356 \approx 4.7551 \] Finally, we need to multiply by \( \cos(5) \) (approximately \( 0.2837 \)): \[ f'(5) \approx 1.2837 - 4.7551 \cdot 0.2837 \] Calculating \( 4.7551 \cdot 0.2837 \): \[ 4.7551 \cdot 0.2837 \approx 1.3491 \] Thus, \[ f'(5) \approx 1.2837 - 1.3491 \approx -0.0654 \] So