Finding Time-Derivative of V(x,y) with Chain Rule

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SUMMARY

The discussion focuses on finding the time-derivative of the expression V(x,y) = ay + x²y² using the chain rule and product rule. The user, Niles, correctly identifies the need to differentiate the term x²y² but seeks clarification on applying the chain rule effectively. The response suggests applying the product rule to the term x²y², emphasizing that x is a function of time (t). This approach is essential for accurately computing the time-derivative.

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Niles
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Homework Statement


Hi all.

I have an expression given by V(x,y) = ay+x2y2, where a is a constant. I wish to find the time-derivative of V(x,y), and this is what I have done:

<br /> \frac{dV}{dt} = a\dot y + \frac{d}{dt}x^2y^2,<br />
where the dot over y represents differentiation w.r.t. time. My problem is the last term, and I wish to use the chain-rule, but I am not sure how to use it. Can you give me a push in the right direction?

Thanks in advance.


Niles.
 
Physics news on Phys.org
Apply the product rule first to x^2.y^2
 
Do you know how to do:
\frac{d}{dt}x^2
(Assuming x is a fuction of t)
?
 

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