Originally Posted by

**Soroban** Hello, Dan!

I think I've got it . . .

$\displaystyle f(x) \:=\:\tan3x$

$\displaystyle f(x+h) - f(x) \;=\;\tan(3[x+h]) - tan(3x) \;=\;\tan(3x + 3h) -\tan(3x)$

. . . $\displaystyle = \;\frac{\tan(3x) + \tan(3h)}{1-\tan(3x)\tan(3h)} - \tan(3x) \;=\; \frac{\tan(3x) + \tan(3h) - \tan(3x) + \tan^2\!(3x)\tan(3h)}{1 -\tan(3x)\tan(3h)}$

. . . $\displaystyle = \;\frac{\tan(3h) + \tan^2(3x)\tan(3h)}{1-\tan(3x)\tan(3h)} \;=\;\frac{\tan(3h)[1 + \tan^2(3x)]}{1-\tan(3x)\tan(3h)} \;=\;\frac{\tan(3h)\sec^2(3x)}{1-\tan(3x)\tan(3h)}$

$\displaystyle \frac{f(x+h)-f(x)}{h} \;=\;\frac{\tan(3h)}{h}\cdot\frac{\sec^2(3x)}{1-\tan(3x)\tan(3h)} \;=\;\frac{\sin(3x)}{h\cos(3x)}\cdot\frac{\sec^2(3 x)}{1-\tan(3x)(\tan(3h)}$

. . . $\displaystyle = \;\frac{3}{\cos(3x)}\cdot\frac{\sin(3x)}{3x}\cdot\ frac{\sec^2(3x)}{1-\tan(3x)\tan(3h)} $

$\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h} \;\;=\;\;\lim_{h\to0}\left[ \frac{3}{\cos(3x)}\cdot\frac{\sin(3x)}{3x}\cdot\fr ac{\sec^2(3x)}{1-\tan(3x)\tan(3h)}\right] $

. . . $\displaystyle = \;\;\frac{3}{\cos(0)}\cdot 1 \cdot\frac{\sec^2(0)}{1 - \tan(3x)\tan(0)} \;\;=\;\;\frac{3}{1}\cdot1\cdot\frac{\sec^2(3x)}{1-0}$

Therefore: .$\displaystyle f'(x) \;=\;3\sec^2(3x)$