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October 1st, 2011, 04:15 PM  #1 
Newbie Joined: Oct 2011 Posts: 6 Thanks: 0  Taking the derivative of a function using the chain rule.
I'm kind of struggling with this problem as it was not explicitly covered in class. 1. f(t)=te^(t^2) This is what I did: f(t)=te^(t^2) z=2t f(z)=te^z f'(t)=f'(z) x z' Use product rule to find derivative of te^z. g(t)=t, h(z)=e^z g'(t)=1, h'(z)=e^z f'(z)=g'h+gh' f'(z)= (1)(e^z)+(t)(e^z) f'(z) x z' = (e^z)(t+1) x (2t) = e^(t^2)(2t^22t) < my answer The book however gives the correct answer as e^(t^2)(12t^2) So where am I going wrong? Thank you for your time. 
October 1st, 2011, 04:39 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Taking the derivative of a function using the chain rule
We are given: Differentiating with respect to t, using the product and chain rules, we find: However, in the spirit of the problem, let's write: thus we have: Now, let and . Thus we have: and (this is where we use the chain rule) Express f(t) as a product. Use the product rule for first order diffentiation: Seriously though, I wouldn't bother with substitutions when the product functions are so simple for a first order differentiation. 

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