Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

So I am trying to find:

$$\frac{d}{dx}\frac{\sec{x}}{1+\tan{x}}$$

And tried doing:

$$\frac{(1+\tan{x})(\tan{x}\times\sec{x})-(\sec^{2}{x})(\sec{x})}{(1+\tan{x})^{2}}$$

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

$$\frac{(1+\tan{x})(\tan{x}\times\sec{x})-(\sec^3{x})}{(1+\tan{x})^{2}}$$

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

$$\frac{\tan{x}\times\sec{x}-\sec^{3}{x}}{1+\tan{x}}$$

Then I got:

$$\frac{\sec{x}\times(\tan{x}-\sec^{2}{x})}{1+\tan{x}}$$

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

Update:

So I tried regrouping:

$$\frac{(1+\tan{x})\tan{x}\times(\sec{x}-(\sec^3{x}))}{(1+\tan{x})^{2}}$$

Factored out a $\sec{x}$:

$$\frac{(1+\tan{x})\tan{x}\times\sec{x}(1-(\sec^2{x}))}{(1+\tan{x})^{2}}$$

Which then gives:

$$\frac{(1+\tan{x})\tan{x}\times\sec{x}\times-\tan^{2}{x}}{(1+\tan{x})^{2}}$$

Which then I said:

$$-\frac{\tan^{3}{x}\times\sec{x}}{(1+\tan{x})}$$

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

share|improve this question
5  
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. –  user667648 Oct 24 '11 at 4:21
2  
@anon - Arturo's point still stands: $\frac{ab+c}{a^2}\not=\frac{b+c}{a}$. –  user5137 Oct 24 '11 at 4:24
1  
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. –  user667648 Oct 24 '11 at 4:25

3 Answers 3

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 $1+\tan^2(x)=\sec^2(x)$.

share|improve this answer
up vote 0 down vote accepted

Decided to use product rule:

$$\frac{d}{dx}\frac{\sec{x}}{(1+\tan{x})}$$

$$\frac{(1+\tan{x})(\tan{x}\times\sec{x})-(\sec^3{x})}{(1+\tan{x})^{2}}$$

I just worked on the top for a while:

$$(1+\frac{\sin{x}}{\cos{x}})(\frac{\sin{x}}{\cos{x}}\times\frac{1}{\cos{x}})-(\frac{1}{\cos{x}^{3}})$$

$$(\frac{\sin{x}}{\cos{x}}+\frac{\sin^2{x}}{\cos^2{x}})(\frac{1}{\cos{x}})-(\frac{1}{\cos{x}^{3}})$$

$$(\frac{\sin{x}}{\cos^2{x}}+\frac{\sin^2{x}}{\cos^3{x}})-(\frac{1}{\cos{x}^{3}})$$

$$\frac{\sin{x}}{\cos^2{x}}+\frac{\sin^2{x}}{\cos^3{x}}-\frac{1}{\cos{x}^{3}}$$

$$\frac{\sin{x}}{\cos^2{x}}+\frac{\sin^2{x}-1}{\cos^3{x}}$$

$$\frac{\sin{x}}{\cos^2{x}}+\frac{-\cos^2{x}}{\cos^3{x}}$$

$$\frac{\sin{x}}{\cos^2{x}}-\frac{1}{\cos{x}}$$

$$\frac{\sin{x}}{\cos^2{x}}-\frac{1}{\cos{x}}*\frac{\cos{x}}{\cos{x}}$$

$$\frac{\sin{x}}{\cos^2{x}}-\frac{\cos{x}}{\cos^2{x}}$$

$$\frac{\sin{x}-cos{x}}{\cos^2{x}}$$

$$\sec^2{x}\times(\sin{x}-\cos{x})$$

$$\sin{x}sec^2{x}-\cos{x}\sec^2{x}$$

$$\sin{x}\times\sec^2{x}-\cos{x}\times\sec^2{x}$$

$$\sin{x}\times\sec^2{x}-\frac{1}{\cos{x}}$$

$$\sin{x}\times\sec^2{x}-\sec{x}$$

$$\sin{x}\times\frac{1}{\cos^2{x}}-\sec{x}$$

$$\frac{\sin{x}}{\cos^2{x}}-\sec{x}$$

$$\frac{\sin{x}}{cos{x}}\frac{1}{cos{x}}-\sec{x}$$

$$\tan{x}\sec{x}-\sec{x}$$

$$(\tan{x}-1)\times\sec{x}$$

Put it all over the original denominator:

$$\frac{(\tan{x}-1)\times\sec{x}}{(1+\tan{x})^{2}}$$

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

share|improve this answer
    
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. –  user667648 Oct 25 '11 at 4:03

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.