Derivative of a derivative with respect to something else

[jaderberg]

Homework Statement

If $$x=e^t$$, find $$\frac{dy}{dx}$$ in terms of $$\frac{dy}{dt}$$ and hence prove $$\frac{d^2y}{dx^2}=e\stackrel{-2t}{}(\frac{d^2y}{dt^2}-\frac{dy}{dt})$$

Homework Equations

The Attempt at a Solution

Well i have done the first bit:

$$x=e^t$$

$$\frac{dx}{dt}=x$$

$$\frac{dy}{dx}=x\stackrel{-1}{}\frac{dy}{dt}$$

Thats fine so now to get the second derivative using product rule:

$$\frac{d^2y}{dx^2}=-x\stackrel{-2}{}$$$$\frac{dy}{dt}+x\stackrel{-1}{}$$$$\frac{d(\frac{dy}{dt})}{dx}$$

now my problem is what is $$\frac{d(\frac{dy}{dt})}{dx}$$?

would greatly appreciate a step by step for that part so i can finally understand this one!

 

[rock.freak667]

consider finding $$\frac{d}{dx}(y)$$

$$\frac{d}{dx}=\frac{d}{dy} \frac{dy}{dx}$$

so now you will have
$$\frac{d}{dx}(y)=\frac{d}{dy}(y) \frac{dy}{dx}=1\frac{dy}{dx}$$

 

[jambaugh]

...now my problem is what is $$\frac{d(\frac{dy}{dt})}{dx}$$?

would greatly appreciate a step by step for that part so i can finally understand this one!

You again invoke chain rule:

$$\frac{dx}{dt}\cdot \frac{d(\frac{dy}{dt})}{dx}= \frac{d(\frac{dy}{dt})}{dt}=\frac{d^2 y}{dt^2}$$

That and a little algebra should get you home.

 

[jaderberg]

im sorry but i still don't understand how to do it...
rockfreak: what happens now when i change y in your expression to $$\frac{dy}{dt}$$?

jam: if that was true then i would get $$\frac{d^2y}{dx^2}=-x\stackrel{-2}{}\frac{dy}{dt}+x\stackrel{-1}{}\frac{d^2y}{dt^2}$$

which would produce the wrong answer...

 

[jaderberg]

ok sorry i think i got it...correct me if I am wrong:

i could write $$\frac{d(\frac{dy}{dt})}{dx}=\frac{dt}{dx}\frac{d(\frac{dy}{dt})}{dt}=x\stackrel{-1}{}\frac{d^2y}{dt^2}$$

thanks guys!

 

[cepheid]

rockfreak: what happens now when i change y in your expression to $$\frac{dy}{dt}$$?

You still follow the same rules...in fact, you could use the convention from Newtonian mechanics that a dot over a quantity represents its time derivative:

$$\dot y = \frac{dy}{dt}$$

To make things less cumbersome. However, rockfreak essentially wrote down that:

$$\frac{d \dot y}{dx} = \frac{d \dot y}{dx}$$

which isn't that useful. Jambaugh had the correct advice

jam: if that was true then i would get $$\frac{d^2y}{dx^2}=-x\stackrel{-2}{}\frac{dy}{dt}+x\stackrel{-1}{}\frac{d^2y}{dt^2}$$
which would produce the wrong answer...

Nope, you would not get that. Check it again. You confused x with t at some point.


Let's go over jambaugh's method for clarity. Using the chain rule, and keeping in mind that the dot denotes differentiation w.r.t. time, we can write:

$$\frac{d}{dx}\left(\frac{dy}{dt}\right) = \frac{d \dot y}{dx}$$

$$= \frac{d \dot y} {dt} \frac{dt}{dx} = \frac{d^2y}{dt^2} \frac{dt}{dx}$$

 

[jaderberg]

yeh cheers i think i got that...pretty simple actually once you know what to look for