Simplfiying a trigonometric polynomial

I was doing a derivative problem for calculus.

The problem reads: $y=(\sec{x}+\tan{x})^5$ find $y'$.

I have found a derivative, I believe is almost certainly correct as I have checked it with a derivative calculator and done it many times.

My result is:
$$5(\tan{x}+\sec{x})^4(\sec{x}\tan{x}+\sec^2{x})$$

I have multiplied the parts together yet for some reason am having trouble getting to the right answer which according to the book is

$5\sec{x}(\sec{x} + \tan{x})^5$

I was wondering how the derivative I found could be simplified down to the correct answer. Unless somebody sees a glaring error with my derivative but like I said I have checked it with many sources. Thank you for your help.

• There is nothing wrong with your answer, and it is actually equivalent to the "right answer". They have simply taken out a fact of $\sec x$ from the last bracket (and what is left makes your 4th power factor into a fifth power), which makes it look somewhat different. Dec 15, 2013 at 0:29

You got the answer right, it is $5\cdot(\tan(x)+\sec(x))^4\cdot(\sec(x)\tan(x)+\sec^2(x))$. You just need to simplify:
$5\cdot(\tan(x)+\sec(x))^4\cdot(\sec(x)(\tan(x)+\sec(x))) = 5\cdot(\tan(x)+\sec(x))^5 \cdot (\sec(x))$