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October 1st, 2011, 04:15 PM   #1
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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.
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October 1st, 2011, 04:39 PM   #2
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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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