Product rule with sqrt(x) and tan(x)

Write $h(x) = f(x)g(x)$ with $f(x) = \sqrt{x}$ and $g(x) = \tan(x)$. Then $f'(x) = \frac{1}{2\sqrt{x}}$ and $g'(x) = \sec^2(x)$. By the product rule, $h'(x) = \frac{1}{2\sqrt{x}}\tan(x) + \sqrt{x}\sec^2(x)$.

Now substitute $x = \pi/4$. We have $\sqrt{\pi/4} = \frac{\sqrt{\pi}}{2}$, $\tan(\pi/4) = 1$, and $\sec^2(\pi/4) = 2$. Therefore, $h'(\pi/4) = \frac{1}{2 \cdot \frac{\sqrt{\pi}}{2}} \cdot 1 + \frac{\sqrt{\pi}}{2} \cdot 2 = \frac{1}{\sqrt{\pi}} + \sqrt{\pi} = \frac{1 + \pi}{\sqrt{\pi}} \approx 2.34$.

The other term is $\frac{1}{2\sqrt{x}}\tan(x)$, which comes from differentiating the square root factor while leaving tangent unchanged, per the product rule. The two terms summed give the full derivative. Recognizing both pieces confirms correct application of the product rule.