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October 9th, 2015, 04:57 PM   #1
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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?
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October 9th, 2015, 10:22 PM   #2
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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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