Question

Here's a question and the answer; how did they get there?
Question: Find dm/dv for m=m_0/sqrt(1-(v^2/c^2)
Answer: dm/dv= (m_0*v)/(c^2(1-(v^2/c^2))^3/2)

Answer 1

\[m=\frac{ m _{o} }{ \sqrt{1-\frac{ v^2 }{ c^2 }} }\] can be written as\[m= m _{o}(1-\frac{ v^2 }{ c^2 })^{-1/2}\]

Answer 2

So does that mean they're assuming the \[v^2/c^2\] term is a constant?

Answer 3

Since this isn't multivariable calculus, I doubt that we're intended to work with two constants. Perhaps their intention is that we treat that term AS the constant?
Does that sound reasonable?

Answer 4

No, mo is constant, v is your variable. Morepresents the initial m value, and that's unchanging and constant, generally speaking.

Answer 5

What do we do with the c^2 term then? Is that a constant also, simply by virtue of not being included as part of the dm/dv notation?

Answer 6

Yes, dm/dv means you're taking the derivative of m with respect to v. Similarly, when you take derivative of y with respect to x, that's just dy/dx. Just treat v like it was your "x"in a sense.

Also,\[\frac{ v^2 }{ c^2 }=\frac{ 1 }{ c^2 }*v^2\]
The reason I wrote it like that was to show how 1/c^2 is just like any other constant in front of your variable just like taking the derivative of 2x^2.

Answer 7

That makes sense. I also just looked and realized I mistyped the answer slightly.

Thank you very much for your help with that!

Answer 8

Cool, no problem.