Question

What is the derivative of sin(1/t)?

Answer 1

Chain rule comes to mind?

Answer 2

Yes, however I'm confused regarding the sin

Answer 3

Refresher:
\[\Large \frac{d}{dt}f[g(t)]= f'[g(t)]g'(t)\]

Answer 4

Yes I understand that. I'm wondering if I would use the quotient rule for (1/t)

Answer 5

You can, or you can also just use the power rule and recall that
\[\Large \frac1t= t^{-1}\]

Answer 6

Ah alright. That's what I always miss. I think I can get the rest. Thanks :D

Answer 7

Good. Post your answers if you can :D

Answer 8

Will do.

Answer 9

So the derivative is cos(1/t)(-1t^-2) right?

Answer 10

That is correct. Just try to write that more neatly (maybe do away with the negative exponent and use a fraction instead) but for the most part, well done ^_^

Answer 11

(in my humble opinion) this looks better:
\[\Large -\frac{\cos\left(\frac1t\right)}{t^2}\]

Answer 12

Yes, that is how my text book would state it. I will get used to doing that :D

Answer 13

Also, I'm doing this for a velocity problem, and a follow up question is when would the velocity be it's greatest. Can you help me out with understanding that part?

Answer 14

v(t) = sin(1/t) was the equation for the velocity and the answer I just gave you was for acceleration.

Answer 15

Here's the full question

Answer 16

A particle moves along the x-axis so that at any time t, 0.1<0.3, its velocity is given by v(t) = sin(1/t).
How would I find when the velocity is at the greatest?