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.