Let \( (x)=7 \cos x+3 \tan x \) \( f^{\prime}(x)=\square \) \( f^{\prime}\left(\frac{11 \pi}{6}\right)=\square \)

To find \( f'(x) \) for the function \( f(x) = 7 \cos x + 3 \tan x \), we will differentiate each term separately. 1. The derivative of \( 7 \cos x \) is: \[ \frac{d}{dx}(7 \cos x) = -7 \sin x \] 2. The derivative of \( 3 \tan x \) is: \[ \frac{d}{dx}(3 \tan x) = 3 \sec^2 x \] Combining these results, we have: \[ f'(x) = -7 \sin x + 3 \sec^2 x \] Now, we need to evaluate \( f'\left(\frac{11 \pi}{6}\right) \). First, we calculate \( \sin\left(\frac{11 \pi}{6}\right) \) and \( \sec^2\left(\frac{11 \pi}{6}\right) \): - The angle \( \frac{11 \pi}{6} \) is in the fourth quadrant, where \( \sin\) is negative. Specifically: \[ \sin\left(\frac{11 \pi}{6}\right) = -\frac{1}{2} \] - The cosine of \( \frac{11 \pi}{6} \) is: \[ \cos\left(\frac{11 \pi}{6}\right) = \frac{\sqrt{3}}{2} \] Therefore, the secant is: \[ \sec\left(\frac{11 \pi}{6}\right) = \frac{1}{\cos\left(\frac{11 \pi}{6}\right)} = \frac{2}{\sqrt{3}} \] Thus, the secant squared is: \[ \sec^2\left(\frac{11 \pi}{6}\right) = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3} \] Now substituting these values into \( f'\left(\frac{11 \pi}{6}\right) \): \[ f'\left(\frac{11 \pi}{6}\right) = -7 \left(-\frac{1}{2}\right) + 3 \left(\frac{4}{3}\right) \] Calculating each term: \[ = \frac{7}{2} + 4 \] Converting \( 4 \) to a fraction with a denominator of 2: \[ = \frac{7}{2} + \frac{8}{2} = \frac{15}{2} \] Thus, we have: \[ f'(x) = -7 \sin x + 3 \sec^2 x \] and \[ f'\left(\frac{11 \pi}{6}\right) = \frac{15}{2} \] Final answers: \[ f'(x) = -7 \sin x + 3 \sec^2 x \] \[ f'\left(\frac{11 \pi}{6}\right) = \frac{15}{2} \]