My Math Forum Taking the derivative of a function using the chain rule.

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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: $f(t)=te^{-t^2}$ Differentiating with respect to t, using the product and chain rules, we find: $f'(t)=t$$e^{-t^2}\(-2t$$\)+(1)$$e^{-t^2}$$=e^{-t^2}$$1-2t^2$$$ However, in the spirit of the problem, let's write: $g(t)=-t^2\:\therefore\:\frac{dg}{dt}=-2t$ thus we have: $f(t)=te^{g(t)}$ Now, let $u(t)=t$ and $v(g)=e^{g}$. Thus we have: $\frac{du}{dt}=1$ and $\frac{dv}{dt}=\frac{dv}{dg}\cdot\frac{dg}{dt}=e^{g }\frac{dg}{dt}=-2te^{-t^2}$ (this is where we use the chain rule) Express f(t) as a product. $f(t)=u(t)v(g)$ Use the product rule for first order diffentiation: $\frac{df}{dt}=u\cdot\frac{dv}{dt}+\frac{du}{dt}\cd ot v=t\cdot-2te^{-t^2}+(1)e^{g(t)}=$ $-2t^2e^{-t^2}+e^{-t^2}=e^{-t^2}$$1-2t^2$$$ Seriously though, I wouldn't bother with substitutions when the product functions are so simple for a first order differentiation.

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# differentiation of te^-2t

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