paritial derivative of function of dependent variablesby ato Tags: dependent, derivative, function, paritial, variables 

#19
Jan513, 08:59 AM

Math
Emeritus
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Thanks
PF Gold
P: 38,904

If F(x,y) is a function of the two variables, x and y, and it is true that y= g(x), then we can write f(x)= F(x, g(x)) and find the derivative of f with the chain rule:
[tex]\frac{df}{dx}= \frac{\partial F}{\partial x}+ \frac{\partial F}{\partial y}\frac{dg}{dx}[/tex] 



#20
Jan513, 12:04 PM

P: 810





#21
Jan813, 02:59 AM

P: 30

lets consider a textbook question
1. ##f(x,y,z) = x^{2} + y^{2} + z^{2}=x^{2}+x^{4}+4x^{2}=x^{4}+5x^{2}## 2. ##\Rightarrow f(x,y,z)=x^{4}+5x^{2}## 3. ##\Rightarrow\frac{\partial}{\partial x}f(x,y,z)=\frac{\partial}{\partial x}(x^{4}+5x^{2})## 4. ##\Rightarrow\frac{\partial}{\partial x}f(x,y,z)= 4x^{3} + 10x## but the author says ##\frac{\partial}{\partial x}f(x,y,z)= 2x## this might sound little annoying, but i still dont know how am i wrong ? 1. ##\frac{df}{dx}=\frac{d}{dx}f(x)## and by extension ##f=f(x)## or 2. ##\frac{df}{dx}\neq\frac{d}{dx}f(x)## what is ##1## since ##1## is a set ? what is ##\{1,2,3\}## ? why make everything set ? p.s. sorry for the late reply. thank you 



#22
Jan813, 04:44 AM

Emeritus
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PF Gold
P: 9,018

\begin{align} \frac{\partial f(x,y,z)}{\partial x}&=D_1f(x,y,z)=(s\mapsto f(s,y,z))'(x)=\lim_{h\to 0}\frac{f(x+h,y,z)f(x,y,z)}{h}\\ &=\lim_{h\to 0}\frac{(x+h)^2x^2}{h}=\frac{d}{dx}x^2.\end{align} The specification ##x=t, y=t^2, z=2t## is a red herring. It's only relevant if you're asked to compute $$\frac{d}{dt}f(x(t),y(t),z(t))=\frac{d}{dt}(5t^2+4t^4).$$ Same thing as = is equal to. ##\frac{df}{dx}## is a bad notation for f'(x). I don't want to write things like f=f(x) since f is a function and f(x) an element of its range. But people sometimes do use these notations interchangeably. It's convenient to make everything sets, because then the theory only needs to leave two things undefined: what a set is, and what it means for a set to be a member of a set. 



#23
Jan813, 07:01 AM

P: 30

##\frac{\partial f(x,y,z)}{\partial t}=\frac{\partial f(x,y,z)}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial f(x,y,z)}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial f(x,y,z)}{\partial z}\frac{\partial z}{\partial t}## and i understood everything except ##\frac{\partial f(x,y,z)}{\partial x},\frac{\partial f(x,y,z)}{\partial y},\frac{\partial f(x,y,z)}{\partial z}##. why should not i take the partial derivative from this equation, ##f(x,y,z)=x^4+5x^2##. whats wrong with that ? you cant ignore anything like that. let ##y=f(x)## so ##\frac{df(x)}{dx}=\frac{dy}{dx}=lim_{h\rightarrow0}\frac{\left.y\right _{x=x}^{x=x+h}}{h}## (what good this (weird) definition of function do anyway ?) thank you 



#24
Jan813, 10:50 AM

Emeritus
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PF Gold
P: 9,018

$$\frac{d}{dt}f(x(t),y(t),z(t))=\frac{d}{dt}(t^4+5t^2)=4t^3+10t.$$ $$f(x,y,z)=x^2+y^2+z^2$$ really means $$\forall x,y,z\in\mathbb R,\quad f(x,y,z)=x^2+y^2+z^2.$$ And this statement is equivalent to $$\forall p,q,r\in\mathbb R,\quad f(p,q,r)=p^2+q^2+r^2.$$ If this is followed by a separate statement x=t, then this can't be the same x, since the first one was a dummy (replaceable) variable and this one isn't. The statements $$x=t, y=t^2, z=2t$$ are intended as definitions of three more functions x,y,z. So what they really mean is this: \begin{align} &\forall t\in\mathbb R,\quad x(t)=t\\ &\forall t\in\mathbb R,\quad y(t)=t^2\\ &\forall t\in\mathbb R,\quad z(t)=2t \end{align} To find ##D_1f##, you only need the definition of f. (Just look at the definition. It doesn't involve any other functions). So these three functions are irrelevant when you compute ##D_1f##. 0=∅ 1={0} 2={0,1} 3={0,1,2} ... I haven't seen a similar definition of the integers in a textbook, but it's not hard to think of one. Let S be any set with two members. Denote those members by p,n. Now consider the set $$Z=\{(n,1),(n,2),\dots\}\cup \{0\}\cup\{(p,1),(p,2),\dots\}.$$ With an appropriate definition of addition, multiplication, etc, the subset $$\{0\}\cup\{(p,1),(p,2),\dots\}$$ will have all the same properties (as far as the standard operations are concerned) as the set {0,1,2,...}. So it has just as much right to be called "the set of positive integers" as {0,1,2,...}. And now we can define the term "integer" as a member of Z. At this point, it would be convenient to simplify the notation like this: Write k instead of (p,k) for all k in {1,2,...}. Write k instead of (n,k) for all k in {1,2,...}. These things are very far from the things you need to understand to compute partial derivatives. 



#25
Jan813, 12:19 PM

P: 30

##3  2 = 1## ##\{ 0,1,2 \}  \{ 0,1 \} = \{ 2 \} \neq 1## is not that a paradox/contradiction. thank you 



#26
Jan813, 12:32 PM

Emeritus
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PF Gold
P: 9,018




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