1. The problem statement, all variables and given/known data The function S_{h}(t) = 30[cos(16.04*)]t models the horizantal position of a pellet with respect to time. Find the first & second derivatives of S_{h}(t). 2. Relevant equations 3. The attempt at a solution I attached a word document because I lack the ability to put together a correctly formatted latex doc in my post. I apologize for the inconvenience. Thank you in advance. Joe 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
Sh(t) = 30[cos(16.04*)]t 1st derivative 30cos(16.04) 2nd 0 If the t is in the cos function, then 1st 16.04*30*(-sin16.04t) 2nd 16.04^2*30*(-cos16.04t)
The first term in the numerator of your limit should be [itex]30(\cos 16.04^\circ)(t+\Delta t)[/itex], so [itex]\Delta t[/itex] gets multiplied by the constant.
I didn't download the paper. I just responded to what the derivatices would be based on what is giving. You need to do the limit definition to obtain the derivatives?
So for my first derivative I get to an answer of delta t/delta t, which I'm sure isn't correct. Are there some rules for differentiating when trig functions are involved that differs from a function say f(x)=x^3 ? Also I posted this question incorrectly as I was going off memory the first time. In the real problem there is no limit written next to the function, does this change things at all? I figured that equation without a limit is just the slope of a secant line. Thanks in advance. Joe
Re: First & Second derivative of a function [Solved] I figured it out. I was missing some rules for derivatives such as f(x)= a constant * a variable, then f'(x) =the constant. Another one was f'(x) of a constant =0. This is of course what Dustin was trying to tell me, I just couldn't put it together from that context. Thanks for your help gentlemen. Joe