Trig Derivative

May 2010
22
0
I have the problem
Find the derivative
\(\displaystyle r=sin^2 3*pi*theta\)

First off, since there are no parentheses in the problem, I don't know if this equals sin(3*pi*theta)^2. I tried that and got a very long answer.
I know the sin will change to cos, but how many of these terms are being operated on by the function of sin?

The book answer was 3pi sin 6pi theta

Thanks.
 
Jan 2010
354
173
So I'm assuming your book writes it like this:

\(\displaystyle r=\sin^2 \, 3 \pi \theta\)

In general it should be interpreted as

\(\displaystyle r=\sin^2 (3 \pi \theta) = \left[ \sin (3 \pi \theta) \right] ^2\)

but I definitely agree that the notation is confusing and should be avoided without parentheses.

Would you mind showing us what answer you got? I should point out that the book is using a trig identity to get to the final answer.

In particular, notice that

\(\displaystyle 2 \sin u \cos u = \sin (2u)\)
 

CaptainBlack

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erehwon
I have the problem
Find the derivative
\(\displaystyle r=sin^2 3*pi*theta\)

First off, since there are no parentheses in the problem, I don't know if this equals sin(3*pi*theta)^2. I tried that and got a very long answer.
I know the sin will change to cos, but how many of these terms are being operated on by the function of sin?

The book answer was 3pi sin 6pi theta

Thanks.
It almost certainly means find:

\(\displaystyle \frac{d}{d\theta} [\sin(3\pi\theta)]^2\)

which should not be much of a problem, you need the chain rule.

The less likely alternative (thought it is what the notation means in other contexts) is:

\(\displaystyle \frac{d}{d\theta} \sin[\sin(3\pi\theta)]\)

which again is a chain rule problem.

CB