# Partial derivatives and the chain rule

• June 18th 2013, 11:46 PM
99.95
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
• June 19th 2013, 02:29 AM
topsquark
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
• June 19th 2013, 04:12 AM
99.95
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)
• June 19th 2013, 05:42 AM
HallsofIvy
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,
$\partial u/\partial x= \partial f/\partial x+ \partial g/\partial x= (\partial f/\partial r)(\partial r/\partial x)+ (\partial g/\partial s)(\partial x)$
(The fact that f depends only on r and g only on s simplifies this. Otherwise we would have $(\partial f/\partial s)(\partial s/\partial x)$ and $(\partial g/\partial r)(\partial r/\partial x)$ as well.)

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

Quote:

But then i am COMPLETELY lost on the third line
Do it again.
$\partial^2 u/\partial x^2= \frac{\partial }{\partial x}(\partial u}{\partial x})= \frac{\partial}{\partial x}(-c\partial f/\partial r+ c\partial g/\partial s)$

$= -c\frac{\partial}{\partial x}(\partial f/\partial r)+ c\partial}{\partial x}(\partial f/\partial s)$
And now use the fact that $\frac{\partial F}{\partial x}= -c\partial F/\partial r+ c\partial F/\partial s$
where F is any differentialble function of x and t. Here, $F= -c\partial f/\partial r+ c\partial f/\partial s$

Taking it one part at a time, $-c\partial (-c \partial f/\partial r+ c\partial f/\partial s)/\partial r= c^2\partial^2f/\partial r^2- c^2\partial f/\partial r\partial s$ and $c\partial(-c\partial f/\partial r+ c\partial f/\partial s)/\partial s= -c^2\partial^2 r/\partial r\partial s+ c^2\partial^2 f\partial s^2$./
Any help would be appreciated.
Please explain to me, like I am an idiot though..
thank MHF, you're always a huge help[/QUOTE]
• June 19th 2013, 05:50 AM
topsquark
Re: Partial derivatives and the chain rule
Not in time!

-Dan
• June 19th 2013, 11:56 PM
99.95
Re: Partial derivatives and the chain rule
Thank you both a million times over!

Makes sense now