Derivative of a partial derivative

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SUMMARY

The discussion centers on calculating the derivative of a partial derivative, specifically the expression for the derivative of U_x with respect to x. The correct formulation is U_xx + U_yx(dy/dx), derived using the chain rule. The participants clarify that U is a function of two variables, x and y, and emphasize the necessity of treating y as a function of x to apply the chain rule effectively. This approach allows for the integration of both partial and total derivatives in the context of multivariable calculus.

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  • Understanding of multivariable calculus concepts, particularly partial derivatives.
  • Familiarity with the chain rule in calculus.
  • Knowledge of differentiable functions and their properties.
  • Ability to manipulate functions of multiple variables.
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  • Explore the concept of mixed partial derivatives, specifically U_yx.
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  • Investigate the relationship between total derivatives and partial derivatives in calculus.
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Students and professionals in mathematics, particularly those studying calculus, as well as educators teaching multivariable calculus concepts.

Chicago_Boy1
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Hello,

So I have the function U(x,y). I have to find a partial derivative of U with respect to x. I understand that one can write that as U subscript x. But now I have to take d/dx of Ux, i.e. I have to take the derivative of Ux(x,y) with respect to x.

Supposedly the answer is Uxx + Uyx(dy/dx), but I don't really understand why...

Note that Ux = U subscript x
Note that Uyx = U subscript yx

Thank you in advance!
 
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Here, in order to be able to talk about df/dx, we have to assume that f is a function of a single variable, x. But to talk about [itex]\partial U/\partial x[/itex] we must have U a function of two variable, x, and, say, y.

To be able to put them together, we must be thinking of y itself as a function of x.

Perhaps it would make more sense to separate the two uses of "x". Think of U as a function of x and y, and then think of x and y as functions of a parameter, t.

If f is any differentiable function of x and y and x and y themselves are differentiable functions of t, by the chain rule,
[tex]\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dx}= f_x\frac{dx}{dt}+ f_y\frac{dy}{dt}[/tex].

But here x= t so that dx/dt= 1 and dy/dt= dx/dt:
[tex]\frac{df}{dx}= \frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}\frac{dy}{dx}= f_x+ f_y\frac{dy}{dx}[/tex]

Replace f by [itex]U_x[/itex] and you have your result.
 

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