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