Treating the derivative notation as a fraction?

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SUMMARY

The discussion centers on the treatment of derivative notation as a fraction in calculus, particularly in the context of solving physics problems involving integrals. Participants clarify that while dy/dx is not technically a quotient, it behaves like one in many scenarios, especially when applying separation of variables in differential equations. Historical perspectives from mathematicians like Leibniz and Euler are referenced, emphasizing that early calculus did not adhere to rigorous definitions, yet the practical application of treating derivatives as fractions yields consistent results. The chain rule is also highlighted as a method to connect derivatives and integrals effectively.

PREREQUISITES
  • Understanding of basic calculus concepts, including derivatives and integrals.
  • Familiarity with the chain rule in calculus.
  • Knowledge of differential equations and separation of variables.
  • Historical context of calculus development by mathematicians like Leibniz and Euler.
NEXT STEPS
  • Study the application of the chain rule in various calculus problems.
  • Explore the concept of separation of variables in solving differential equations.
  • Review the historical development of calculus and its foundational figures.
  • Investigate the rigorous definitions of derivatives as established by Riemann, Cauchy, and Weierstrass.
USEFUL FOR

Students of calculus, physics enthusiasts, and educators seeking to deepen their understanding of derivative notation and its applications in solving integrals and differential equations.

A.MHF
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Homework Statement


I was working on a physics problem that involves integrals and I stumbled upon this:

m*g*sin θ − k*x = m*dv/dt =m*v*dv/dx
→ (m*g*sin θ − k*x)/m =v*dv/dx
→∫[(m*g*sin θ − k*x)/m]dx =∫v*dv

Notice that we multiplied by dx on both sides and dx has been canceled from the right one.
I have learned before and it has been emphasised on me that the derivative notation isn't a fraction and has nothing to do with it, yet sometimes we treat it like one to solve a problem.
How do I make sense of this? Does it make a mathematical sense?

Homework Equations


m*g*sin θ − k*x = m*dv/dt =m*v*dv/dx
→ (m*g*sin θ − k*x)/m =v*dv/dx
→∫[(m*g*sin θ − k*x)/m]dx =∫v*dv

The Attempt at a Solution


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Historically speaking the quantities like dx were treated by Leibniz, Euler, and others as real numbers; very small, non zero numbers. These pioneers did not formulate calculus on the most rigorous of footings, but this was eventually done by other giants like Riemann, Cauchy, and Weierstrass. Yes, you are correct that dy/dx is not technically a quotient. However, the when solving differential equations with separation of variables one does treat dy/dx as a quotient. It is easy to show that the result is the same regardless of whether this technique is used or not (see http://www.math-cs.gordon.edu/courses/ma225/handouts/sepvar.pdf).

So... No, dy/dx is not a quotient, but it behaves like one in many different circumstances.
 
A.MHF said:

Homework Statement


I was working on a physics problem that involves integrals and I stumbled upon this:

m*g*sin θ − k*x = m*dv/dt =m*v*dv/dx
→ (m*g*sin θ − k*x)/m =v*dv/dx
→∫[(m*g*sin θ − k*x)/m]dx =∫v*dv

Notice that we multiplied by dx on both sides and dx has been canceled from the right one.
I have learned before and it has been emphasised on me that the derivative notation isn't a fraction and has nothing to do with it, yet sometimes we treat it like one to solve a problem.
How do I make sense of this? Does it make a mathematical sense?

-
Think of this as using the chain rule.
dv/dt = (dv/dx)⋅(dx/dt) = v⋅(dv/dx), because dx/dt = v

If v is a differentiable function of x, then the differential dx is given by dx = (dv/dx)⋅dx .

Then ##\displaystyle\ \int \left(v\frac{dv}{dx}\right)dx=\int v\,dv\ ##
 

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