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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. Tags chain, derivative, rule Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post JORGEMAL Calculus 1 November 30th, 2013 12:57 PM daigo Calculus 4 July 13th, 2012 06:19 PM google Calculus 1 October 1st, 2011 04:39 PM Peter1107 Calculus 1 September 8th, 2011 10:25 AM dmxnemesis Calculus 31 October 11th, 2010 10:23 PM

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