Having dificulty wth this problems, can someone hep 1. z=(3t+1)(5t+2)
L Lenier New member Joined Feb 19, 2007 Messages 2 Feb 19, 2007 #1 Having dificulty wth this problems, can someone hep 1. z=(3t+1)(5t+2)
pka Elite Member Joined Jan 29, 2005 Messages 11,976 Feb 19, 2007 #2 With what? What are you to do with that? We need more to work with than that.
L Lenier New member Joined Feb 19, 2007 Messages 2 Feb 19, 2007 #3 Find the derivative, using the product rule
pka Elite Member Joined Jan 29, 2005 Messages 11,976 Feb 19, 2007 #4 \(\displaystyle \L\begin{array}{rcl} z(t) = f(t)g(t)\quad & \Rightarrow & \quad z'(t) = f'(t)g(t) + f(t)g'(t) \\ f(t) = \left( {3t + 1} \right)\quad & \Rightarrow & \quad f'(t) = 3 \\ g(t) = \left( {5t + 2} \right)\quad & \Rightarrow & \quad g'(t) = 5 \\ \end{array}\)
\(\displaystyle \L\begin{array}{rcl} z(t) = f(t)g(t)\quad & \Rightarrow & \quad z'(t) = f'(t)g(t) + f(t)g'(t) \\ f(t) = \left( {3t + 1} \right)\quad & \Rightarrow & \quad f'(t) = 3 \\ g(t) = \left( {5t + 2} \right)\quad & \Rightarrow & \quad g'(t) = 5 \\ \end{array}\)