# Partial derivative chain rule

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

where z=z(u)

u=x+at

## Answers and Replies

Homework Helper
Gold Member
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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iScience
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?

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##.

iScience
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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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.

iScience
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?

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.