If \( f(x)=\frac{3 x^{2} \tan x}{\sec x} \), find \( f^{\prime}(x)= \) Find \( f^{\prime}(5)= \)

To find the derivative \( f'(x) \) of the function \( f(x) = \frac{3x^2 \tan x}{\sec x} \), we can first simplify the function. Recall that \( \sec x = \frac{1}{\cos x} \), so we can rewrite \( f(x) \) as: \[ f(x) = 3x^2 \tan x \cos x \] Now, we will 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: - \( u(x) = 3x^2 \) - \( v(x) = \tan x \cos x \) We will need to differentiate both \( u(x) \) and \( v(x) \). 1. Differentiate \( u(x) \): \[ u'(x) = 6x \] 2. Differentiate \( v(x) = \tan x \cos x \) using the product rule again: 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 derivative of \( f(x) \): \[ f'(x) = u'(x)v(x) + u(x)v'(x) \] Substituting \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \): \[ f'(x) = 6x \tan x \cos x + 3x^2 \left( \sec^2 x \cos x - \frac{\sin^2 x}{\cos x} \right) \] Now, we can simplify \( f'(x) \): \[ f'(x) = 6x \tan x \cos x + 3x^2 \sec^2 x \cos x - 3x^2 \frac{\sin^2 x}{\cos x} \] Now, we can evaluate \( f'(5) \). We need to compute \( f'(5) \): 1. Calculate \( \tan(5) \), \( \sec(5) \), and \( \cos(5) \). 2. Substitute these values into the expression for \( f'(x) \). Using a calculator: - \( \tan(5) \approx -0.9589 \) - \( \sec(5) \approx 1.1674 \) - \( \cos(5) \approx 0.2837 \) Now substituting these values into \( f'(5) \): \[ f'(5) = 6(5)(-0.9589)(0.2837) + 3(5^2)(1.1674^2)(0.2837) - 3(5^2)\frac{(-0.9589^2)}{0.2837} \] Calculating each term: 1. First term: \( 6(5)(-0.9589)(0.2837) \approx -8.134 \) 2. Second term: \( 3(25)(1.1674^2)(0.2837) \approx 3(25)(1.3673)(0.2837) \approx 3(25)(0.3875) \approx 28.125 \) 3. Third term: \( 3(25)\frac{(-0.9589^2)}{0.2837} \approx 3(25)(3.0005) \approx 225.0375 \) Now summing these