# How to best simplify a chain/product rule with lots of trig functions?

I've found the derivative of the following:

$$g(x) = \sec(8x)\tan(5x^9)$$

to be

$$g'(x) = 8\sec(8x)\tan(8x)\tan(5x^9) + 45x^8 \sec(8x)(\sec(5x^9))^2$$

I'm aware that the trig identities are interchangeable to an extent, so tan(8x) might be written as sin(8x)/cos(8x). However, I'm not sure if such rules would help simplifying this problem. Once I've got all these trig functions in here, is there any point to fooling around with the identities to try and condense it?

Also, I notice that sec(8x) appears on both sides -- can this be consolidated into (2*sec(8x))? Or, for that matter, take out the 2 and multiply by the 8 to get a 16 in front?

• You can factor out $\sec(8x)$, not add them. – Namaste Feb 28 '15 at 23:21
• Oh, whoops! You're totally right. Score one for simplification! Thanks! – barney Feb 28 '15 at 23:31

Factoring out a common factor $\sec(8x)$: \begin{align} g'(x) & = 8\sec(8x)\tan(8x)\tan(5x^9) + 45x^8 \sec(8x)(\sec(5x^9))^2\\ \\ &=\sec(8x)\Big(8\tan(8x)\tan(5x^9) + 45x^8\sec^2(5x^9)\Big)\\ \\ \end{align}