# Differentiating sin functions

• March 27th 2011, 12:57 PM
Shout
Differentiating sin functions
Not sure how to differentiate these functions, if you could show the steps for either function that'd be great :).

Sorry I also don't know how to format it correctly so bear with me.

Question:
$
p(t) = 70sin((2pi/10)t) +120

$

at t = 6

and

$
P(t) = 3sin(theta -(pi/2))
$

at theta = pi/3

Thanks!
• March 27th 2011, 01:02 PM
pickslides
$(\sin nt)' = n\cos nt$

Also

$(\sin (\theta +n))' = \cos \theta$ by the chain rule.
• March 27th 2011, 03:36 PM
Shout
I can't seem to make the second one work.
This is what I did;
3sin(x-pi/6) becomes
3cos(x-pi/6)*(1)

Whats wrong with that?

The answer is supposed to be 0.05 but i keep getting 2.99.
• March 27th 2011, 04:01 PM
pickslides
By the chain rule

$y= \sin (\theta +n)$

Make $u=\theta +n\implies y = \sin u$

Now find $\displaystyle \frac{dy}{d\theta} = \frac{du}{d\theta}\times \frac{dy}{du}$
• March 27th 2011, 04:06 PM
Shout
I'm still confused, would you be able to show me the next step and then maybe I'll understand?
• March 27th 2011, 04:09 PM
pickslides
It seems your attempt is fine. In post #2 I should've written $\cos(\theta +n)$

$\displaystyle \frac{dy}{du} = \cos u$

$\displaystyle \frac{du}{d\theta} = 1$
• March 27th 2011, 04:16 PM
Shout
Alright, so then it will be y' = 3*cos(pi/3-pi/6) which leaves me at 2.59