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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.

  • 0
    @anon: Your algebra is still wrong. $(1+\tan x)(\tan x\sec x) - \sec^3x \neq (1+\tan x)\tan x(\sec x - \sec^3x).$ If you can't do the algebra right, you have no hope of getting the derivative right.2011-10-24

3 Answers 3

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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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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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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...

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    @ArturoMagidin: :0 Omg. I must be retarded. Consider my mind blown.2011-10-25