Derivative of a quotient involving sec and csc

Using the quotient rule with numerator $u(x) = \sec x + \csc x$ and denominator $v(x) = x$, we have $g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}$. Since $u'(x) = \sec x \tan x - \csc x \cot x$ and $v'(x) = 1$, this gives $g'(x) = \frac{x(\sec x \tan x - \csc x \cot x) - (\sec x + \csc x)}{x^2}.$

At $x = \pi/3$: $\sec(\pi/3) = 2$, $\tan(\pi/3) = \sqrt{3}$, $\csc(\pi/3) = \frac{2\sqrt{3}}{3}$, and $\cot(\pi/3) = \frac{\sqrt{3}}{3}$. So $\sec x \tan x = 2\sqrt{3}$ and $\csc x \cot x = \frac{2\sqrt{3}}{3} \cdot \frac{\sqrt{3}}{3} = \frac{2 \cdot 3}{9} = \frac{2}{3}$.

Substitute into the formula: numerator $= \frac{\pi}{3}\left(2\sqrt{3} - \frac{2}{3}\right) - \left(2 + \frac{2\sqrt{3}}{3}\right)$, and denominator $= \left(\frac{\pi}{3}\right)^2 = \frac{\pi^2}{9}$. Therefore, $g'\left(\frac{\pi}{3}\right) = \frac{\frac{\pi}{3}\left(2\sqrt{3} - \frac{2}{3}\right) - \left(2 + \frac{2\sqrt{3}}{3}\right)}{\frac{\pi^2}{9}}.$

Numerically: $\frac{\pi}{3} \approx 1.047198$, so $\frac{\pi}{3}(2\sqrt{3} - 2/3) \approx 1.047198 \times (3.464102 - 0.666667) \approx 2.930827$. Also $2 + \frac{2\sqrt{3}}{3} \approx 2 + 1.154701 = 3.154701$. The numerator is approximately $2.930827 - 3.154701 = -0.223874$. The denominator is $\frac{\pi^2}{9} \approx 1.096761$. Thus, $g'(\pi/3) \approx \frac{-0.223874}{1.096761} \approx -0.204$.