Chain Rule

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?

jamesbob said:
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?

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

Thanks very much

Hurkyl
Staff Emeritus