1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ You can factor out $\sec(8x)$, not add them. $\endgroup$ – Namaste Feb 28 '15 at 23:21
  • $\begingroup$ Oh, whoops! You're totally right. Score one for simplification! Thanks! $\endgroup$ – barney Feb 28 '15 at 23:31
1
$\begingroup$

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

I wouldn't worry about trying to condense further; trying too hard to beautify answers, especially on timed tests, can eat up your time and prevent you from completing the assignment or test,

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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