# Treating the derivative notation as a fraction?

1. Apr 12, 2015

### A.MHF

1. The problem statement, all variables and given/known data
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?
2. Relevant 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
3. The attempt at a solution
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2. Apr 12, 2015

### brainpushups

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. Apr 12, 2015

### SammyS

Staff Emeritus
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\$