# Calculus Problem of the Week November 4, 2011

Sometimes easy and sometimes hard, our calculus problem of the week could come from any calculus topic. If you really want to get better at calculus, following these problems is a great way to make yourself practice! Past calculus problems of the week.

This week’s problem:
(click “see the solution” at the bottom of post to, well, see the solution. If it doesn’t work, please click the title of this post and then try (I’m working on this!))

As compared to the last couple of weeks, this week’s problem is more of an exercise than a “problem”. Perhaps you will see what I mean!

Find the derivative of $f(x)=4(\cos x - \sin x)^2 + 8 \sin x \cos x$.

See the solution.

While I think random exercises to practice even the “easier” concepts are always a good idea, I did leave a little trick in this one.

Before you take the derivative, see if you can find any way to simplify things to a more useful form. For example, if you expand the squared term :

$f(x)=4(\cos x - \sin x)^2 + 8 \sin x \cos x$
=$f(x)=4(\cos^2 x - 2\sin x \cos x + \sin^2 x) + 8 \sin x \cos x$
=$f(x)= 4( 1 - 2\sin x \cos x) + 8 \sin x \cos x$
=$f(x)= 4( 1 - 2\sin x \cos x) + 8 \sin x \cos x = 4$

Yes this function is just 4, so the derivative is 0. Even if you did this without finding all the terms that cancel, you should have found the same answer here, but it is way more fun to find a shortcut! They aren’t always there but watch for terms that may cancel or simplify before starting any derivative or integral. (ok SOMETIMES this backfires as you need the extra terms, but this is rare!)