Partial derivatives and the chain rule

Im revising for an exam i have next week, and cannot for my life understand the solutions to this previous exam question:

http://i.imgur.com/hLGMwIp.jpg

Is this essentially the same question as:

z = f(x,y) , g = g(s,t), y = h(s,t)

and then find dz/ds and dz/dt ?

the solutions look like the employ implicit differentiation, but i am not sure. typically I am confused from the second line,

HOw is du/dr * dr/dt = -c * df/dr ?

My train of thought is dr/dt = -c

du/dr is another way of saying df/dr, and using that same deduction we get the second half of that equation

But then i am COMPLETELY lost on the third line

Any help would be appreciated.

Please explain to me, like I am an idiot though..

thank MHF, you're always a huge help

Re: Partial derivatives and the chain rule

Quote:

Originally Posted by

**99.95** du/dr is another way of saying df/dr

Point of clarification: how is this line true?

-Dan

Re: Partial derivatives and the chain rule

I was just taking a guess, assuming that the change in notation allowed them to get the solution.

Not sure what they are doing.(Wondering)

Re: Partial derivatives and the chain rule

Quote:

Originally Posted by

**99.95** Im revising for an exam i have next week, and cannot for my life understand the solutions to this previous exam question:

http://i.imgur.com/hLGMwIp.jpg
Is this essentially the same question as:

z = f(x,y) , g = g(s,t), y = h(s,t)

Typo: you mean x= g(s, t), right?

Quote:

and then find dz/ds and dz/dt ?

the solutions look like the employ implicit differentiation, but i am not sure. typically I am confused from the second line,

HOw is du/dr * dr/dt = -c * df/dr ?

Yes, that's the chain rule. You are given that u(x, t)= f(x- ct)+ g(x+ ct) and they are setting r= x- ct, s(x+ ct) so that u(x, t)= f(r)+ g(s).

Quote:

My train of thought is dr/dt = -c

Yes, that is true.

Quote:

du/dr is another way of saying df/dr, and using that same deduction we get the second half of that equation

That's rather *roughly* what is happening. More specifically, since u= f(r)+ g(s) and r and s are functions or x and t,

$\displaystyle \partial u/\partial x= \partial f/\partial x+ \partial g/\partial x= (\partial f/\partial r)(\partial r/\partial x)+ (\partial g/\partial s)(\partial x)$

(The fact that f depends **only** on r and g **only** on s simplifies this. Otherwise we would have $\displaystyle (\partial f/\partial s)(\partial s/\partial x)$ and $\displaystyle (\partial g/\partial r)(\partial r/\partial x)$ as well.)

$\displaystyle = -c \partial f/\partial r)+ c\partial g/\partial s$

Quote:

But then i am COMPLETELY lost on the third line

Do it **again**.

$\displaystyle \partial^2 u/\partial x^2= \frac{\partial }{\partial x}(\partial u}{\partial x})= \frac{\partial}{\partial x}(-c\partial f/\partial r+ c\partial g/\partial s)$

$\displaystyle = -c\frac{\partial}{\partial x}(\partial f/\partial r)+ c\partial}{\partial x}(\partial f/\partial s)$

And now use the fact that $\displaystyle \frac{\partial F}{\partial x}= -c\partial F/\partial r+ c\partial F/\partial s$

where F is **any** differentialble function of x and t. Here, $\displaystyle F= -c\partial f/\partial r+ c\partial f/\partial s$

Taking it one part at a time, $\displaystyle -c\partial (-c \partial f/\partial r+ c\partial f/\partial s)/\partial r= c^2\partial^2f/\partial r^2- c^2\partial f/\partial r\partial s$ and $\displaystyle c\partial(-c\partial f/\partial r+ c\partial f/\partial s)/\partial s= -c^2\partial^2 r/\partial r\partial s+ c^2\partial^2 f\partial s^2$./

Any help would be appreciated.

Please explain to me, like I am an idiot though..

thank MHF, you're always a huge help[/QUOTE]

Re: Partial derivatives and the chain rule

Re: Partial derivatives and the chain rule

Thank you both a million times over!

Makes sense now