from the chain rule we have

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from the chain rule we have

[math]

h(x) = f(g(x))

h(x)' = f(g(x))'g(x)'

\frac{\partial h(x))}{\partial x}=\frac{\partial f(g(x))}{\partial g(x)}\frac{\partial g(x)}{\partial x}

[/math]

but is the "expanded form" like so, for semantic understanding only of course.

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>>7803609

[math]

h(x) = f(g(x))

\newline

h(x)' = f(g(x))'g(x)'

\newline

\frac{\partial h(x))}{\partial x}=\frac{\partial f(g(x))}{\partial g(x)}\frac{\partial g(x)}{\partial x}

[/math]

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>>7803609

[math]h(x) = f(g(x)) \newline h(x)' = f(g(x))'g(x)' \newline [/math]

HOLY FUCK

[math]

\frac{\partial h(x))}{\partial x}=\frac{\partial f(g(x))}{\partial g(x)}\frac{\partial g(x)}{\partial x}[/math]

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>>7803614

please

[math]

h(x) = f(g(x))[/math]

[math] h(x)' = f(g(x))'g(x)'

[/math][math] \frac{\partial h(x))}{\partial x}=\frac{\partial f(g(x))}{\partial g(x)}\frac{\partial g(x)}{\partial x}[/math]

>>

[math]

h(x) = f(g(x))

[/math]

[math]

h(x)' = f(g(x))'g(x)'

[/math]

[math]

\frac{\partial h(x))}{\partial x}=\frac{\partial f(g(x))}{\partial g(x)}\frac{\partial g(x)}{\partial x}

[/math]

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>>7803616

the last one should be [math]\frac{\partial h}{\partial x} (x) = \frac{\partial f}{\partial x} (g(x)) \frac{\partial g}{\partial x} (x)[/math]

you have to differentiate first, and take the derivative at the appropriate point

>>

What is your point?

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>>7803625

so number 2 is correct?

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>>7803629

yes

f'(g(x)) means "the derivative of f taken at the point g(x)".

derivative of f is df/dx

derivative of f taken at the point g(x) is the function df/dx evaluated at the point g(x).

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>>7803632

is this also correct then? if so how come g is at the bottom this time

>>

These are not partial derivatives and you should not be using that notation, since it leads to confusion.

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>>7803642

can you please explain or atleast link me, Im really lost :c

>pic related, my solid foundation in calculus

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>>7803645

all he means is that when you're referring to a derivative of a single variable function you just use a d

you're using the symbol for partial differentiation, as though f and g are multivariable functions

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>>7803640

plz halp with this

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>>7803645

Well I'm sure you can find a derivation of the chain rule in many many places.

Partial differentiation implies that the function you are differentiating is implicitly dependent on the variable. I don't honestly know if it's the same in all cases and I don't plan to figure that out, but it can lead to confusion when interpreting the process, which I think is why you made this thread.

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>>7803640

that's when f is a function of g.

For example

f(t) = t^2 (f is a function of time).

but it turns out t= x/3 (t is a function of position).

But you can't just write df/dx easily, since f is not a function of x.

you can write df/dt * dt/dx

I think I see why you are confused.

you should really use f' to denote the derivative of a function of 1 variable.

There is no ambiguity. if f is the square function, f'(x) = 2x, f'(3t) = 2*(3t).

if you want to introduce a new variable, you need the chain rule.

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>>7803669

I cant differentiate on how those two are different,

cant I view it like this

f(g)

h(x) = f(x^2)

so then g is a function of x

idun understand anything anymore :C

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>>7803640

if you were to use that notation, I think it's necessary to include a second term:

[math]

\partial_g f \cdot \partial_x g + \partial_x f

[/math]

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>>7803691

you're right, just please stop using partial derivatives. For the love of all that is the false god.

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>>7803700

because he's an idiot. just do some example problems both ways and figure that shit out.

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