Question

If $f(x) = \sec x - \cot x$, then $f'\!\left(\dfrac{\pi}{6}\right)$ equals

• A. $-3$
• B. $\dfrac{14}{3}$
• C. $\dfrac{11}{3}$
• D. 5

Hint:

$\dfrac{d}{dx}(\sec x) = \sec x\tan x$; $\dfrac{d}{dx}(\cot x) = -\csc^2 x$. Note the sign: $f = \sec x - \cot x$, so $f' = \sec x\tan x + \csc^2 x$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Differentiate using standard derivatives of $\sec x$ and $\cot x$.
Step 2: Detailed Explanation:
$f'(x) = \sec x\tan x + \csc^2 x$.
At $x=\pi/6$: $\sec(\pi/6)=\dfrac{2}{\sqrt{3}}$, $\tan(\pi/6)=\dfrac{1}{\sqrt{3}}$, $\csc(\pi/6)=2$.
$f'(\pi/6) = \dfrac{2}{\sqrt{3}}\cdot\dfrac{1}{\sqrt{3}} + 4 = \dfrac{2}{3} + 4 = \dfrac{14}{3}$.
Step 3: Final Answer:
$f'(\pi/6) = \dfrac{14}{3}$.