# Partial derivative chain rule

could someone show me how $\frac{∂}{∂t}$($\frac{∂z}{∂u}$)= $\frac{∂^2z}{∂u^2}$ $\frac{∂u}{∂t}$

where z=z(u)

u=x+at

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LCKurtz
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could someone show me how $\frac{∂}{∂t}$($\frac{∂z}{∂u}$)= $\frac{∂^2z}{∂u^2}$ $\frac{∂u}{∂t}$

where z=z(u)

u=x+at
##z## is only a function of ##u## so you might as well use ##z'(u)## instead of partials. I will denote ##\frac{\partial z}{\partial u}## as ##z'(u)##. Your equation then becomes$$\frac \partial {\partial t}z'(u) = \frac {d}{du}z'(u)\frac{\partial u}{\partial t} =z''(u)\frac{\partial u}{\partial t}$$Isn't that exactly the chain rule on ##z'(u)##?

 Corrected some TeX partials vs d/dt symbols.

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i apologize, but can you please show me what you did step by step? i know how to apply the chain rule for functions when the function itself is actually given, ie the derivative of the function times the derivation of the inside. but i don't understand what we're doing now.

where did you get the d/du in the front from? and the ∂u/∂t that's behind z', that's just from the derivative of the inside right?

LCKurtz
Homework Helper
Gold Member
i apologize, but can you please show me what you did step by step? i know how to apply the chain rule for functions when the function itself is actually given, ie the derivative of the function times the derivation of the inside. but i don't understand what we're doing now.

where did you get the d/du in the front from? and the ∂u/∂t that's behind z', that's just from the derivative of the inside right?
If ##f## is a function of ##u## and ##u## is a function of ##x## and ##y## then ##f## is ultimately a function of ##x## and ##y##. It makes sense to ask for the partials of ##f## with respect to ##x## and ##y##. The corresponding chain rules are (I will use subscripts for partials to save typing): $$f_x = f'(u)u_x,~f_y = f'(u)u_y$$Note the use of the ' on ##f## since ##f## itself is a function of just one variable and the partial symbols on ##u## since it is a function of more than one variable. These formulas are almost certainly in your text.

Your problem is solved by applying one of these chain rules to ##z'(u)## with second variables ##x## and ##t## instead of ##x## and ##y##.

i understand z'(t)=$\frac{∂z}{∂u}$$\frac{∂u}{∂t}$ and z'(x)=$\frac{∂z}{∂u}$$\frac{∂u}{∂x}$

i just don't understand the leap from that to this: $\frac{∂}{∂t}$$\frac{∂z}{∂u}$

Your problem is solved by applying one of these chain rules to z′(u) with second variables x and t instead of x and y.
but 'u' is the intermediate variable. i don't know what to do with $\frac{∂z}{∂u}$. if it involved finding the partial of z with respect to x and t, i could just apply the chain rule like you say, but it's asking me to find the partial of z with respect to u, the intermediate variable. what can i do with that? i can't use x and y.

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LCKurtz
Homework Helper
Gold Member
i understand z'(t)=$\frac{∂z}{∂u}$$\frac{∂u}{∂t}$ and z'(x)=$\frac{∂z}{∂u}$$\frac{∂u}{∂x}$

i just don't understand the leap from that to this: $\frac{∂}{∂t}$$\frac{∂z}{∂u}$

but 'u' is the intermediate variable. i don't know what to do with $\frac{∂z}{∂u}$. if it involved finding the partial of z with respect to x and t, i could just apply the chain rule like you say, but it's asking me to find the partial of z with respect to u, the intermediate variable. what can i do with that? i can't use x and y.
It becomes exactly what I showed you in post #2.

okay can you tell me if i did this correctly

$\frac{∂}{∂t}$($\frac{∂z}{∂u}$)=$\frac{∂}{∂u}$($\frac{∂z}{∂t}$)=$\frac{∂}{∂u}$($\frac{∂z}{∂u}$$\frac{∂u}{∂t}$)

i basically just swapped the positions of the u and t in the beginning is this right?

if i swap the u and the t, aren't i applying clairaut's theorem? and if i am, then don't i need to prove that zut is continuous and ztu is continuous since that's the condition for which clairaut's theorem applies?

LCKurtz
Homework Helper
Gold Member
okay can you tell me if i did this correctly

$\frac{∂}{∂t}$($\frac{∂z}{∂u}$)=$\frac{∂}{∂u}$($\frac{∂z}{∂t}$)=$\frac{∂}{∂u}$($\frac{∂z}{∂u}$$\frac{∂u}{∂t}$)

i basically just swapped the positions of the u and t in the beginning is this right?

if i swap the u and the t, aren't i applying clairaut's theorem? and if i am, then don't i need to prove that zut is continuous and ztu is continuous since that's the condition for which clairaut's theorem applies?
That is not the setting for Clairaut's theorem. I don't know why you are doing this. The answer to your question in your post #1 is that it is the chain rule I explained in post #4. That is where the formula comes from, and the proof of the chain rule is probably in your text.