Question

Please help!: Calculus problem
Let f (x) = sin(x) cot(x).
(a) Use the product rule to
nd f 0(x).
(b) True or false: for all real numbers x, f (x) = cos(x).
(c) Explain why the function that you found in (a) is almost the opposite of the
sine function, but not quite. (Hint: convert all of the trigonometric functions
in (a) to sines and cosines, and work to simplify. Think carefully about the
domain of f and the domain of f 0.)
So I know how to do a, and I believe that b is true.. I'm stuck with c though, what is it really asking? I simplified f'(x) and used the basic trig functions.

Answer 1

b is false since by definition:
\[f(x) = sin(x) cot(x) = \sin(x) \frac{\cos(x)}{\sin(x)} \]
Note that this \(f(x)\) is undefined when \(\sin(x) = 0\) so we can ONLY say:
\[ f(x) = cos(x) \quad \iff \sin(x) \neq 0\]

Answer 2

This should help you explain c better, which isn't talking about the derivatives, just \(f(x)\)

Answer 3

yes, I realized that b is false just while you were typing a reply.
It said, "the function you found in a" wasn't it the derivative function?

Answer 4

I see, I misread. Note that, since \(f(x) = \cos(x)\) for \(x \neq 2n\pi\) then:

\( f'(x) = -\sin(x) \) for \(x \neq 2n\pi\) since it wasn't in the domain of \(f(x)\).
This is just the "opposite" of sine, except it is undefined at certain points for reasons above.