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from the chain rule we have
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from the chain rule we have
$h(x) = f(g(x)) h(x)' = f(g(x))'g(x)' \frac{\partial h(x))}{\partial x}=\frac{\partial f(g(x))}{\partial g(x)}\frac{\partial g(x)}{\partial x}$

but is the "expanded form" like so, for semantic understanding only of course.
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>>7803609
$h(x) = f(g(x)) \newline h(x)' = f(g(x))'g(x)' \newline \frac{\partial h(x))}{\partial x}=\frac{\partial f(g(x))}{\partial g(x)}\frac{\partial g(x)}{\partial x}$
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>>7803609
$h(x) = f(g(x)) \newline h(x)' = f(g(x))'g(x)' \newline$

HOLY FUCK
$\frac{\partial h(x))}{\partial x}=\frac{\partial f(g(x))}{\partial g(x)}\frac{\partial g(x)}{\partial x}$
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>>7803614
$h(x) = f(g(x))$
$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}$
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$h(x) = f(g(x))$
$h(x)' = f(g(x))'g(x)'$
$\frac{\partial h(x))}{\partial x}=\frac{\partial f(g(x))}{\partial g(x)}\frac{\partial g(x)}{\partial x}$
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>>7803616
the last one should be $\frac{\partial h}{\partial x} (x) = \frac{\partial f}{\partial x} (g(x)) \frac{\partial g}{\partial x} (x)$

you have to differentiate first, and take the derivative at the appropriate 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
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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:
$\partial_g f \cdot \partial_x g + \partial_x f$
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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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>>7803697
>>7803625

so how come he wants me to do this crazy shit
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>>7803700
because he's an idiot. just do some example problems both ways and figure that shit out.