Chain Rule Confusion

[laser1]

Homework Statement

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Relevant Equations

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Source: Paul's Notes (https://tutorial.math.lamar.edu/Solutions/CalcIII/ChainRule/Prob6.aspx)

In his solutions:

where he computes the stuff in red. My question is, why is $$\frac{\partial y}{\partial p}$$not $$1+3\frac{\partial t}{\partial p}=1+3\left(\frac{-1}{2}\right)=\frac{-1}{2}$$ Thanks

 

[andrewkirk]

Because the rule for partial differentiation is to calculate the derivative using the formula given and not looking through any variables that may themselves be functions of other variables. In your alternative approach you are treating t as a function of p, looking through that function definition and including as part of the partial derivative any dependencies on p that you find . You are not supposed to look through like that. Treat each variable in the formula given for y as if it is an atomic variable that has no dependence on any other variable.
Unfortunately, the curly-d notation for partial differentiation tends to obscure this important principle because it makes it look like partial differentiation is something you do to a variable. It isn't. It's something you do to a function. Total differentiation can be something we do to a variable, but not partial differentiation.
See my Insights not on this topic, which may help clarify: https://www.physicsforums.com/insights/partial-differentiation-without-tears/

 

[laser1]

Because the rule for partial differentiation is to calculate the derivative using the formula given and not looking through any variables that may themselves be functions of other variables. In your alternative approach you are treating t as a function of p, looking through that function definition and including as part of the partial derivative any dependencies on p that you find . You are not supposed to look through like that. Treat each variable in the formula given for y as if it is an atomic variable that has no dependence on any other variable.
Unfortunately, the curly-d notation for partial differentiation tends to obscure this important principle because it makes it look like partial differentiation is something you do to a variable. It isn't. It's something you do to a function. Total differentiation can be something we do to a variable, but not partial differentiation.
See my Insights not on this topic, which may help clarify: https://www.physicsforums.com/insights/partial-differentiation-without-tears/

Thanks, I read through your article. My flaw there was computing dy/dp as opposed to partial y / partial p.

One point is still annoying me though - what if we consider $p$ to be a placeholder for $1-2t$? So $y=(1-2t)+3t-4s=1+t-4s$. This is the exact same thing, right? But now, $$\frac{\partial w}{\partial t}=\text{TERMS}+\frac{\partial w}{\partial y} \frac{\partial y}{\partial t}$$ where TERMS is just the other partial derivative terms. Now the part that goes through $y$ is equal to $1$.

I can't really see where the flaw is with this logic.

 

[PeroK]

Thanks, I read through your article. My flaw there was computing dy/dp as opposed to partial y / partial p.

One point is still annoying me though - what if we consider $p$ to be a placeholder for $1-2t$? So $y=(1-2t)+3t-4s=1+t-4s$. This is the exact same thing, right? But now, $$\frac{\partial w}{\partial t}=\text{TERMS}+\frac{\partial w}{\partial y} \frac{\partial y}{\partial t}$$ where TERMS is just the other partial derivative terms. Now the part that goes through $y$ is equal to $1$.

I can't really see where the flaw is with this logic.

If you simplify first, and eliminate $p$, then the second and third terms in the full expansion reduce to a single term involving $\frac{\partial y}{\partial t}$.

 

[laser1]

If you simplify first, and eliminate $p$, then the second and third terms in the full expansion reduce to a single term involving $\frac{\partial y}{\partial t}$.

I mean, if $y$ was defined as $y=(1-2t)+3t-4s=1+t-4s$, that's exactly what we'd do, right? We would only have those two terms in the expression involving $\partial w/ \partial y$, those terms being $\frac{\partial w}{\partial y} \frac{\partial y}{\partial t}$.

 

[PeroK]

I mean, if $y$ was defined as $y=(1-2t)+3t-4s=1+t-4s$, that's exactly what we'd do, right? We would only have those two terms in the expression involving $\partial w/ \partial y$.

Yes. If you do it both ways, you must get the same final answer. If you eliminate $p$ first, then the whole expansion is simplified.

 

[laser1]

Yes. If you do it both ways, you must get the same final answer. If you eliminate $p$ first, then the whole expansion is simplified.

 

[PeroK]

I mean, if $y$ was defined as $y=(1-2t)+3t-4s=1+t-4s$, that's exactly what we'd do, right?

Note that, IMO, that is how $y$ is defined. It is that function of $s$ and $t$. That is not changed by the fact that we have an intermediate variable $p$.

 

[PeroK]

View attachment 353774

Nonsense. Do the whole calculation for $\frac{\partial w}{\partial t}$.

 

[laser1]

Ah my bad! I completely missed that the solutions had the $\frac{\partial w}{\partial y} \frac{\partial y}{\partial t}$ term next to it, which evaluates to $3\frac{\partial w}{\partial y}$, and $3-2=1$ so it all works out the same!