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So I am trying to find:


And tried doing:


Because of the Quotient Rule. Then I did some simplifying:


Further simplification (crossed out the $(1+\tan{x})^{2})$:


Then I got:


But Wolfram Alpha says differently. Where did I go wrong? Thanks.


So I tried regrouping:


Factored out a $\sec{x}$:


Which then gives:


Which then I said:


Which still isn't right. Sorry, if I made another obvious mistake.

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The error is in "Further simplification". You cannot cancel like that. $$\frac{ab+c}{a^2}$$ is not equal to $\frac{b+c}{a}$. So you cannot cancel $(1+\tan x)$ as you did. – Arturo Magidin Oct 24 '11 at 4:18
@Aturo: Sorry I kind of wrote it wrong the second time. I edited it. – Dair Oct 24 '11 at 4:21
@anon - Arturo's point still stands: $\frac{ab+c}{a^2}\not=\frac{b+c}{a}$. – user5137 Oct 24 '11 at 4:24
It's still wrong. The first part of the denominator has a $1+\tan$ factor to cancel but the second part, that is $-\sec^3$, doesn't. Also, have fun sharing my inbox :) – anon Oct 24 '11 at 4:24
@Aturo: O wait, I see what you are getting at... Ok, I'll try again and see how it goes. – Dair Oct 24 '11 at 4:25

When you crossed out the $1+\tan(x)$, you left the $\sec^3(x)$ unchanged, which you can't do.

Instead of doing that, try expanding the product in the numerator, and using the identity $1+\tan^2(x)=\sec^2(x)$.

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up vote 0 down vote accepted

Decided to use product rule:



I just worked on the top for a while:





















Put it all over the original denominator:


So that is what I did, there probably should be an easier way though...

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From $\sin^2 x + \cos^2 x = 1$, dividing through by $\cos^2 x$ you get $\tan^2x + 1 = \sec^2 x$. So you can replace $\sec^3 x$ by $$\sec^3x = sec^2x\sec x = (1+\tan^2x)\sec x.$$Then you have$$\frac{(1+\tan x)\sec x - \sec^3x}{(1+\tan x)^2} = \frac{(1+\tan x)\sec x - (\tan^2x - 1)\sec x}{(1+\tan x)^2}.$$Now write $(\tan^2 x - 1) = (\tan x +1)(\tan x -1)$, and now you can factor $(\tan x + 1)$ from the numerator and cancel. – Arturo Magidin Oct 25 '11 at 3:02
@ArturoMagidin: :0 Omg. I must be retarded. Consider my mind blown. – Dair Oct 25 '11 at 4:03

Here is a straightforward way of finding the derivative manipulating only sines and cosines:

$$ \frac{d}{dx} \left[ \frac{\sec(x)}{1+\tan(x)} \right] \\ = \frac{d}{dx} \left[ \frac{1}{\cos(x)} \frac{1}{1+\frac{\sin(x)}{\cos(x)}}\right] \\ = \frac{d}{dx} \left[ \frac{1}{\cos(x)+\sin(x)}\right] \\ = \frac{d}{dx} \left[ (\cos(x) + \sin(x))^{-1} \right] \\ = (-1)(\cos(x) + \sin(x))^{-2}(-\sin(x) + \cos(x)) \\ = \frac{(-1)(-\sin(x)+\cos(x))}{(\cos(x)+\sin(x))(\cos(x)+\sin(x))} \\ =\frac{\sin(x) - \cos(x)}{\cos^2(x)+2\sin(x)\cos(x) + \sin^2(x)} \\ = \frac{\sin(x)-\cos(x)}{1+2\sin(x)\cos(x)} \\ = \frac{\sin(x) -\cos(x)}{1+\sin(2x)}$$

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