# Thread: Derivative of tan(3x) the long way

1. ## Derivative of tan(3x) the long way

I had a student ask this the other day and I could find no immediate way to solve it without doing some serious work on it. Since then I've been mulling it over and came up with the following course of action to simplify the work. I'd like y'all to take a look at it and tell me if I've got it right.

$\lim_{h \to 0}\frac{tan(3[x + h]) - tan(3x)}{h}$

Let y = 3x. Then the limit is
$\lim_{h \to 0}\frac{tan(y + 3h) - tan(y)}{h}$

Now let H = 3h. Then the limit is
$\lim_{H \to 0}\frac{tan(y + H) - tan(y)}{\frac{H}{3}}$

$= 3~\lim_{H \to 0}\frac{tan(y + H) - tan(y)}{H}$

Then we do our magic with taking the derivative of the tangent function and get
$= 3~sec^2(y) = 3~sec^2(3x)$

Is this process of substitution acceptable in general? It works for this problem, but I'd like to verify that this can be generalized.

Thanks!
-Dan

2. Hello, Dan!

I think I've got it . . .

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

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

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

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

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

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

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

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

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

3. Originally Posted by Soroban
Hello, Dan!

I think I've got it . . .

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

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

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

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

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

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

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

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

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

I knew it could be done in something like this manner. (Nice job, by the way!) I was looking for some kind of simple shortcut as this was a High School aged Calculus student. I figured that (s)he wouldn't be able to handle the full treatment. (Though I readily admit that I didn't have the details to do it this way to mind.)

I was looking more for an comment whether or not my "shortcut" method is applicable to any derivative. I realize that this is the basis for one specific application of the chain rule, but I'm not good enough at the "picky details" of Calculus to know if it's general.

-Dan

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