# Chain Rule

1. Mar 13, 2006

### jamesbob

Question: Let
$$Q = \sqrt{x^2 + y}e^t$$​
where (for t > or = 0)
$$x = \sqrt{1 - e^{-2t}}$$​
and
$$y = 2 - e^{-2t}$$​

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:

$$Q = \sqrt{1 - e^{-2t} + 2 - e^{-2t}}e^t = \sqrt{3 - 2e^{-2t}}e^t$$

Now applying the product and chain rule:

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

$$\frac{du}{dz} = \frac{1}{2\sqrt{z}}, \left \frac{dz}{dt} = 4e^-2t$$

So $$\frac{du}{dt} = \frac{4e^{-2t}}{2\sqrt{3 - 2e^{-2t}}}$$

As for v, $$v = e^t, \left \frac{dv}{dt} = e^t$$

So $$\frac{dQ}{dt} = \sqrt{3 - 2e^{-2t}}e^t + \frac{4e^{-t}}{2\sqrt{3 - 2e^{-2t}}}$$

Is this right so far and can it be simplified further?

2. Mar 13, 2006

### topsquark

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.

-Dan

3. Mar 14, 2006

### jamesbob

Thanks very much

4. Mar 14, 2006

### Hurkyl

Staff Emeritus
While a perfectly valid technique, I think part of the point of this problem is not to do such a substitution.

5. Mar 14, 2006

### jamesbob

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?

6. Mar 14, 2006

### Hurkyl

Staff Emeritus
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!