Understanding the Partial Derivative Chain Rule for z=z(u) and u=x+at

[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

 

[LCKurtz]

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 $z'(u)$?

[Edit] Corrected some TeX partials vs d/dt symbols.

 

[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?

 

[LCKurtz]

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.

 

[LCKurtz]

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 z_ut is continuous and z_tu is continuous since that's the condition for which clairaut's theorem applies?

 

[LCKurtz]

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 z_ut is continuous and z_tu 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.