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[tex] Q = \sqrt{x^2 + y}e^t[/tex]where (for t > or = 0)

[tex] x = \sqrt{1 - e^{-2t}} [/tex]and

[tex] y = 2 - e^{-2t} [/tex]

Usingthe 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?

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# Homework Help: Chain Rule

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