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

To find the derivative \( f'(x) \) of the function \( f(x) = \frac{2x^2 \tan x}{\sec x} \), we can first simplify the function. Recall that \( \sec x = \frac{1}{\cos x} \), so we can rewrite \( f(x) \) as: \[ f(x) = 2x^2 \tan x \cos x \] Now, we can differentiate \( f(x) \) using the product rule. The product rule states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then: \[ (uv)' = u'v + uv' \] In our case, we can let: - \( u(x) = 2x^2 \) - \( v(x) = \tan x \cos x \) We will need to differentiate both \( u(x) \) and \( v(x) \). 1. Differentiate \( u(x) \): \[ u'(x) = \frac{d}{dx}(2x^2) = 4x \] 2. Differentiate \( v(x) = \tan x \cos x \) using the product rule: Let \( a(x) = \tan x \) and \( b(x) = \cos x \). - \( a'(x) = \sec^2 x \) - \( b'(x) = -\sin x \) Using the product rule: \[ v'(x) = a'(x)b(x) + a(x)b'(x) = \sec^2 x \cos x + \tan x (-\sin x) \] Now, we can simplify \( v'(x) \): \[ v'(x) = \sec^2 x \cos x - \tan x \sin x \] Since \( \tan x = \frac{\sin x}{\cos x} \), we can rewrite \( \tan x \sin x \) as: \[ \tan x \sin x = \frac{\sin^2 x}{\cos x} \] Thus, we have: \[ v'(x) = \sec^2 x \cos x - \frac{\sin^2 x}{\cos x} \] Now, substituting back into the product rule for \( f(x) \): \[ f'(x) = u'(x)v(x) + u(x)v'(x) \] Substituting \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \): \[ f'(x) = 4x \tan x \cos x + 2x^2 \left( \sec^2 x \cos x - \frac{\sin^2 x}{\cos x} \right) \] Now, we can simplify \( f'(x) \): \[ f'(x) = 4x \tan x \cos x + 2x^2 \sec^2 x \cos x - 2x^2 \frac{\sin^2 x}{\cos x} \] Now, we can evaluate \( f'(4) \). First, we need to calculate \( \tan(4) \), \( \sec(4) \), and \( \cos(4) \). Using a calculator or software, we find: - \( \tan(4) \approx 1.1578 \) - \( \sec(4) \approx 1.0039 \) - \( \cos(4) \approx -0.6536 \) Now substituting \( x = 4 \) into \( f'(x) \): 1. Calculate \( 4 \tan(4) \cos(4) \): \[ 4 \cdot 1.1578 \cdot (-0.6536) \approx -3.036 \] 2. Calculate \( 2 \cdot 4^2 \sec^2(4) \cos(4) \): \[ 2 \cdot 16 \cdot (1.0039^2) \cdot (-0.6536) \approx 2 \cdot 16 \cdot 1.0078 \cdot (-0.6536) \approx -21.052 \] 3. Calculate \( 2 \cdot 4^2 \frac{\sin^2(4)}{\cos(4)} \): Using \( \sin(4) \approx -0.7568 \): \[ 2 \cdot 16 \cdot \frac{(-0.7568)^2}{-0.6536} \approx 32 \cd