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 October 9th, 2015, 04:57 PM #1 Newbie   Joined: Oct 2015 From: Wisconsin Posts: 1 Thanks: 0 Second derivative and chain rule I've been trying to solve this problem but I'm just so confused and stumped: I need to find the SECOND derivative of q = 3sin(t - 1)/ π The first derivative I found was: 3 cos (t-1) / π * (1 / π ) My question here is. I know that π can't be 0 because that would be dividing by 0. But does anyone know why that is? Now for the second I'm stumped: I know the derivative of cos is -sin but what do i do with the 3? Doesn't it just turn into a 0? Would everything be exactly the same except for cos?
 October 9th, 2015, 10:22 PM #2 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 From your first derivative, I'm assuming you mean $$q = 3\sin\left(\dfrac{t - 1}{\pi}\right).$$ Now, you're right in saying $$\dfrac{dq}{dt} = \dfrac{3}{\pi}\cos\left(\dfrac{t - 1}{\pi}\right),$$ but $\pi$ is a constant ($\pi = 3.14159...$) so it can never be zero. For the second derivative, just treat the $\dfrac{3}{\pi}$ the same way you treated the $3$ in the first derivative.

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