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 gives us: \[ f'(x) = \sec x - \frac{(\tan x - 4) \tan x}{\sec x} \] Now, we can express \( \frac{1}{\sec x} \) as \( \cos x \): \[ f'(x) = \sec x - (\tan x - 4) \tan x \cos x \] Now, we need to evaluate \( f'(5) \). First, we need to 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 \( f'(x) \): \[ f'(5) = \sec(5) - (\tan(5) - 4) \tan(5) \cos(5) \] Calculating \( \cos(5) \): - \( \cos(5) \) is approximately \( 0.2837 \) Now substituting: \[ f'(5) = 1.2837 - (-0.9589 - 4)(-0.9589)(0.2837) \] Calculating \( -0.9589 - 4 = -4.9589 \): \[ f'(5) = 1.2837 - (-4.9589)(-0.9589)(0.2837) \] Calculating \( (-4.9589)(-0.9589)(0.2837) \): \[ = 4.9589 \cdot 0.9589 \cdot 0.2837 \approx 1.339 \] Finally, we have: \[ f'(5) \approx 1.2837 - 1.339 \approx -0.0553 \] Thus, the final answers are: \[ f'(x) = \sec x - \frac{(\tan x - 4) \tan x}{\sec x} \] \[ f'(5) \approx -0.0553 \]