- #1

SchroedingersLion

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Hello there,

I have stumbled across further examples to derivatives of multivariable functions that confuse me. Similar to my other thread:

https://www.physicsforums.com/threads/partial-derivative-of-composition.985371/#post-6309196

Suppose we have two functions, ## f: R^2 \rightarrow R, (t,x) \mapsto f(t, x) ## and ##g: R \rightarrow R, t \mapsto g(t)## .

We have $$\frac {df} {dt} = \frac {\partial f} {\partial x}\frac {\partial x} {\partial t} + \frac {\partial f} {\partial t} .$$

If we now write ##x=g(t)## and consider ##f(t, g(t))## what do people actually mean by this?

Then ##\frac {\partial } {\partial t} f(t, g(t)) = \frac {\partial f(t, x)} {\partial t}##, and ## \frac {d} {dt} f(t, g(t))= \frac {\partial f} {\partial x}\frac {\partial x} {\partial t} + \frac {\partial f} {\partial t}## as above.

So seeing that the total derivatives are equal for both cases, the interpretation decides the outcome of the partial derivative. I would have guessed that Option B is actually "correct". In a simpler case: If I have ##f(x)## and ##x(t)## (all simple 1D functions of the reals), I would write $$\frac {\partial } {\partial t} f(x)=0 \\ \frac {d} {dt} f(x) = \frac {\partial f} {\partial x} \frac {\partial x} {\partial t} \\ \text{and } \frac {\partial } {\partial t} f(x(t)) = \frac {\partial } {\partial t} (f \circ x)(t) = \frac {\partial x} {\partial t} \frac {\partial } {\partial x} f(x) = \frac {d} {dt} f(x) $$

So I would always assume ##f(x(t))## to imply a composition. Yet I have seen authors that treated it still as ##f(x)##. Is there something wrong in my understanding, or is there really room for ambiguity here?

Wikipedia seems to contradict itself:

" In this case, we are actually interested in the behavior of the composite function ##f(x, y(x))## . The partial derivative of ##f## with respect to

If they viewed it as a composite function, then the partial derivative of that composite function

https://en.wikipedia.org/wiki/Total_derivative#Example:_Differentiation_with_direct_dependencies

I have stumbled across further examples to derivatives of multivariable functions that confuse me. Similar to my other thread:

https://www.physicsforums.com/threads/partial-derivative-of-composition.985371/#post-6309196

Suppose we have two functions, ## f: R^2 \rightarrow R, (t,x) \mapsto f(t, x) ## and ##g: R \rightarrow R, t \mapsto g(t)## .

We have $$\frac {df} {dt} = \frac {\partial f} {\partial x}\frac {\partial x} {\partial t} + \frac {\partial f} {\partial t} .$$

If we now write ##x=g(t)## and consider ##f(t, g(t))## what do people actually mean by this?

**Option A)**I still view it as the function ##f##, simply evaluated at ##x=g(t)##.Then ##\frac {\partial } {\partial t} f(t, g(t)) = \frac {\partial f(t, x)} {\partial t}##, and ## \frac {d} {dt} f(t, g(t))= \frac {\partial f} {\partial x}\frac {\partial x} {\partial t} + \frac {\partial f} {\partial t}## as above.

**Option B)**I take ##f(t, g(t))## to be a partial composition, i.e. I have a new function ##h(t)=f(t,g(t))##. In that case, partial and total time derivatives are equal and should also be equal to the total time derivative of the interpretation of A).So seeing that the total derivatives are equal for both cases, the interpretation decides the outcome of the partial derivative. I would have guessed that Option B is actually "correct". In a simpler case: If I have ##f(x)## and ##x(t)## (all simple 1D functions of the reals), I would write $$\frac {\partial } {\partial t} f(x)=0 \\ \frac {d} {dt} f(x) = \frac {\partial f} {\partial x} \frac {\partial x} {\partial t} \\ \text{and } \frac {\partial } {\partial t} f(x(t)) = \frac {\partial } {\partial t} (f \circ x)(t) = \frac {\partial x} {\partial t} \frac {\partial } {\partial x} f(x) = \frac {d} {dt} f(x) $$

So I would always assume ##f(x(t))## to imply a composition. Yet I have seen authors that treated it still as ##f(x)##. Is there something wrong in my understanding, or is there really room for ambiguity here?

**edit:**Wikipedia seems to contradict itself:

" In this case, we are actually interested in the behavior of the composite function ##f(x, y(x))## . The partial derivative of ##f## with respect to

*##x##*does not give the true rate of change of ##f## with respect to changing ##x## because changing ##x## necessarily changes*##y##.*"If they viewed it as a composite function, then the partial derivative of that composite function

**should**give the whole variation...https://en.wikipedia.org/wiki/Total_derivative#Example:_Differentiation_with_direct_dependencies

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