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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^2-2t) <-- my answer The book however gives the correct answer as e^(-t^2)(1-2t^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.  Tags chain, derivative, function, rule, taking ,
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# differentiation of te^-2t

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