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.
>>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]
>>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]
>>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]
>>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?
>>7803625
so number 2 is correct?
>>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).
>>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.
>>7803642
can you please explain or atleast link me, Im really lost :c
>pic related, my solid foundation in calculus
>>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
>>7803640
plz halp with this
>>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.
>>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.
>>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
>>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]
>>7803691
you're right, just please stop using partial derivatives. For the love of all that is the false god.
>>7803700
because he's an idiot. just do some example problems both ways and figure that shit out.
