Undergrad Calculate partial derivatives and mixed partial derivatives

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When calculating partial derivatives of a function with three variables, such as f(x,y,z), differentiating with respect to one variable (e.g., ∂x) while holding the others constant does not imply that the result is a function of x only. The partial derivative ∂x f(x,y,z) still depends on y and z, even if they are treated as constants during differentiation. It is incorrect to express ∂x f(x,y,z) as a function solely of x, such as g(x), unless y and z are eliminated from the equation. Evaluating the derivative at a specific point allows for a function in one variable, but this is a different consideration. Understanding this distinction clarifies the nature of partial derivatives in multivariable calculus.
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Hi. I know how to calculate partial derivatives and mixed partial derivatives such as ∂2f/∂x∂y but I've now become confused about something. If I have a function of 3 variables eg. f(x,y,z) and I calculate ∂x then I am differentiating wrt x while holding y and z constant. Does that mean ∂x then becomes a function of x only ie does ∂x f(x,y,z) = φ( x ) ? If it does then ∂y and ∂z will always be zero but I know this is not the case. I'm confused !
 
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dyn said:
Hi. I know how to calculate partial derivatives and mixed partial derivatives such as ∂2f/∂x∂y but I've now become confused about something. If I have a function of 3 variables eg. f(x,y,z) and I calculate ∂x then I am differentiating wrt x while holding y and z constant. Does that mean ∂x then becomes a function of x only ie does ∂x f(x,y,z) = φ( x ) ? If it does then ∂y and ∂z will always be zero but I know this is not the case. I'm confused !
No. It's still a function of three variables.
Differentiating along a single variable simply means to consider the change in values along this coordinate.

You might want to play around a little with Wolfram, e.g. http://www.wolframalpha.com/input/?i=f(x,y)+=+xy^3++4+x^2
Imagine the partial derivation ##\partial_x## as a tangent in ##x-##direction. It still varies with ##y## and ##z##.
 
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Thanks. So it is incorrect to write ∂x f(x, y ,z) = g ( x ) ?
 
dyn said:
Thanks. So it is incorrect to write ∂x f(x, y ,z) = g ( x ) ?
Unless the variables y and z disappear as a result of differentiating wrt x, they are still there. They might disappear, but don't count on it.
 
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dyn said:
Thanks. So it is incorrect to write ∂x f(x, y ,z) = g ( x ) ?
Yes.

You could of course evaluate the derivative at some point ##p=(x_0,y_0,z_0)## and get
$$\frac{\partial}{\partial_x}\bigg{|}_p f(x,y,z) = g(x_0,y_0,z_0)$$
Or if you want to examine the ##x-##component, you could consider ##g(x,y_0,z_0)## and get a function in one variable, because you fixed ##y=y_0## and ##z=z_0##. But this is another issue.
 
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Relativistic Momentum, Mass, and Energy Momentum and mass (...), the classic equations for conserving momentum and energy are not adequate for the analysis of high-speed collisions. (...) The momentum of a particle moving with velocity ##v## is given by $$p=\cfrac{mv}{\sqrt{1-(v^2/c^2)}}\qquad{R-10}$$ ENERGY In relativistic mechanics, as in classic mechanics, the net force on a particle is equal to the time rate of change of the momentum of the particle. Considering one-dimensional...

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