How to Isolate dx/dt from d(xy)/dt and dy/dt?

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    Differentiation
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SUMMARY

The discussion focuses on isolating the derivative \(\frac{dx}{dt}\) from the expression \(\frac{d(xy)}{dt}\) and \(\frac{dy}{dt}\) using the product rule in calculus. The correct formula derived is \(\frac{dx}{dt} = \frac{1}{y} \left( \frac{d(xy)}{dt} - x \frac{dy}{dt} \right)\). The user is implementing this in an ODE solver in MATLAB, indicating that the issue may lie elsewhere in the code rather than in the mathematical derivation itself.

PREREQUISITES
  • Understanding of calculus, specifically the product rule for differentiation.
  • Familiarity with ordinary differential equations (ODE) and their solvers.
  • Proficiency in MATLAB programming and syntax.
  • Knowledge of derivatives and their applications in mathematical modeling.
NEXT STEPS
  • Review the product rule in calculus to reinforce understanding of differentiation techniques.
  • Explore MATLAB's ODE solver functions, such as ode45, for better implementation.
  • Study examples of isolating variables in differential equations to enhance problem-solving skills.
  • Learn about debugging techniques in MATLAB to identify and fix code errors effectively.
USEFUL FOR

This discussion is beneficial for students and professionals in mathematics, engineering, and computer science, particularly those working with calculus and MATLAB for solving differential equations.

Timeforheroes0
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Hey guys,
Just a simple question to see if I'm going insane. So say I have an expression for:
\frac{dxy}{dt} and an expression for \frac{dy}{dt}, how to I go about isolating an expression for just \frac{dx}{dt}
I have tried using the product rule but it doesn't work for what I want. I know this is a very simple question but I haven't done calculus in a long time!
 
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<br /> \frac{d(xy)}{dt}=\frac{dx}{dt}y+x\frac{dy}{dt}\Rightarrow \frac{dx}{dt}=\frac{1}{y}(\frac{d(xy)}{dt}-x\frac{dy}{dt})<br />
 
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Yeah that's what I did.. I'm doing this in an ode solver in MATLAB so obviously my code is wrong elsewhere.
thanks!
 

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