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Homework Help: Chain Rule

  1. Mar 13, 2006 #1
    Question: Let
    [tex] Q = \sqrt{x^2 + y}e^t[/tex]​
    where (for t > or = 0)
    [tex] x = \sqrt{1 - e^{-2t}} [/tex]​
    [tex] y = 2 - e^{-2t} [/tex]​

    Using the chain rule calculate dQ/dt, expressing your answer in as simple a form as possible.

    My work so far

    Subbing in values of x and y:

    [tex] Q = \sqrt{1 - e^{-2t} + 2 - e^{-2t}}e^t = \sqrt{3 - 2e^{-2t}}e^t [/tex]

    Now applying the product and chain rule:

    [tex] u = (3 - 2e^{-2t})^{\frac{1}{2}} \left chain \left rule \Rightarrow u = z^{\frac{1}{2}} \left z = 3 - 2e^-2t [/tex]

    [tex] \frac{du}{dz} = \frac{1}{2\sqrt{z}}, \left \frac{dz}{dt} = 4e^-2t [/tex]

    So [tex] \frac{du}{dt} = \frac{4e^{-2t}}{2\sqrt{3 - 2e^{-2t}}} [/tex]

    As for v, [tex] v = e^t, \left \frac{dv}{dt} = e^t [/tex]

    So [tex] \frac{dQ}{dt} = \sqrt{3 - 2e^{-2t}}e^t + \frac{4e^{-t}}{2\sqrt{3 - 2e^{-2t}}} [/tex]

    Is this right so far and can it be simplified further?
  2. jcsd
  3. Mar 13, 2006 #2
    Looks good to me. Theoretically you would want to remove the square root from the denominator and add the two expressions, but I personally don't see that that actually simplifies anything.

  4. Mar 14, 2006 #3
    Thanks very much :smile:
  5. Mar 14, 2006 #4


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    While a perfectly valid technique, I think part of the point of this problem is not to do such a substitution.
  6. Mar 14, 2006 #5
    Hmm, i see, so would i just do the chian rule keeping x and y in, and then sub values in right at the end?
  7. Mar 14, 2006 #6


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    Staff Emeritus
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    Or maybe not even sub in the values at all -- the final result might (or might not) look nicer in terms of x and y!
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