Treating the derivative notation as a fraction?

In summary, the conversation discusses the concept of treating the derivative notation as a quotient and whether it makes mathematical sense. The solution involves using the chain rule and integrating v⋅(dv/dx) to get ∫v⋅dv. However, it is important to note that while this technique may be used to solve differential equations, the derivative notation is not technically a quotient.
  • #1
A.MHF
26
1

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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  • #2
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.
 
  • #3
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?

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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\ ##
 

FAQ: Treating the derivative notation as a fraction?

What does it mean to treat the derivative notation as a fraction?

When we say to treat the derivative notation as a fraction, we are essentially rewriting the derivative expression using fraction notation instead of the traditional notation. This allows us to apply algebraic rules and properties to simplify the expression and make it easier to solve.

Why is it helpful to treat the derivative notation as a fraction?

Treating the derivative notation as a fraction can be helpful because it allows us to use familiar algebraic rules and properties to solve the expression. This can make the process of finding the derivative easier and more efficient.

Can we always treat the derivative notation as a fraction?

Yes, we can always treat the derivative notation as a fraction. The derivative of a function can be thought of as the slope of the tangent line at a given point, and we can use the slope formula (rise over run) to rewrite the derivative expression as a fraction.

Are there any limitations to treating the derivative notation as a fraction?

One limitation to treating the derivative notation as a fraction is that it may not always result in a simplified expression. In some cases, it may be more efficient to use other methods, such as the power rule or the chain rule, to find the derivative.

Can treating the derivative notation as a fraction help us understand the concept better?

Yes, treating the derivative notation as a fraction can help us gain a better understanding of the concept. By using familiar algebraic rules and properties, we can see how the derivative relates to the original function and how changes in the input affect the output.

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