Find the derivative of $y=\tan x,secx$ and evaluate it at $x=\frac{\pi }{4}$.

Step-by-step solution

1. Apply the product rule: $\frac{d}{dx}[\tan x,secx]=(\frac{d}{dx}\tan x)secx+\tan x,\frac{d}{dx}secx$.
2. Use $\frac{d}{dx}\tan x=se{c}^{2}x$ and $\frac{d}{dx}secx=secx,\tan x$.
3. Substitute: $y^{\prime }=se{c}^{2}x,secx+\tan x,secx,\tan x=se{c}^{3}x+secx,{\tan }^{2}x$.
4. Factor $secx$: $y^{\prime }=secx(se{c}^{2}x+{\tan }^{2}x)$.
5. Use $se{c}^{2}x=1+{\tan }^{2}x$ to get $se{c}^{2}x+{\tan }^{2}x=1+2{\tan }^{2}x$. Hence $y^{\prime }=secx(1+2{\tan }^{2}x)$.
6. Evaluate at $x=\frac{\pi }{4}$: $sec!\left(\frac{\pi }{4}\right)=\sqrt{2}$ and $\tan !\left(\frac{\pi }{4}\right)=1$.
7. So $y^{\prime }!\left(\frac{\pi }{4}\right)=\sqrt{2},(1+2\cdot 1)=\sqrt{2}\cdot 3=3\sqrt{2}$.

Concept tested: Derivative of product of trig functions

Published 1 October 2026 · How this solution was prepared and checked