Partial derivatives chain rule

  • Thread starter sid9221
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  • #1
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Suppose we have a function [tex] V(x,y)=x^2 + axy + y^2[/tex]
how do we write
[tex] \frac{dV}{dt} [/tex]


For instance if [tex] V(x,y)=x^2 + y^2[/tex], then [tex] \frac{dV}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} [/tex]

So, is the solution

[tex] \frac{dV}{dt} = 2x \frac{dx}{dt} + ay\frac{dx}{dt} + ax\frac{dy}{dt} + 2y \frac{dy}{dt} [/tex]
 

Answers and Replies

  • #3
ChrisVer
Gold Member
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Just for generality, whenever you have a function [itex]F(x(t),y(t),...,z(t), t)[/itex] (it is a function of some functions of t, and also depends explicitly on t, for example:
[itex]F= x(t)+y(t)^{2}+...+lnz(t) + (t^{3}-t^{2})[/itex]
and you want to find its derivative, you have:
[itex]\frac{dF}{dt}= \frac{\partial F}{\partial x}|_{y,...,z=const}\frac{dx}{dt}+\frac{\partial F}{\partial y}|_{x,...,z=const}\frac{dy}{dt}+...+\frac{\partial F}{\partial z}|_{x,y,...=const}\frac{dz}{dt} + \frac{\partial F}{\partial t}|_{x,y,...,z=const}[/itex]
 

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