What kind of operations are allowed on derivatives?

[madah12]

Homework Statement

I just started in calculus ii ,and I remember that most I didn't use specific rules when dealing with derivatives and I sometimes managed to get away with it(especially in physics. now I have been playing with relations and I got unacceptable results so i will post what I did ( even if all of it is wrong)
let y=f(x)
y''=dy'/dx=dy'/dy dy/dx
=y' dy'/dy
y''/y'=dy'/dy
1/y' *dy'/dx = dy'/dy
1/y' dy' = dy'/dy *dx
1/y' dy' = d/dy * dy/dx *dx
and I canceled the dy and the dx
and got
ln(y')=1
I know this is stupid but how many operations I did were wrong?

Homework Equations

The Attempt at a Solution

 

[rock.freak667]

Homework Statement

I just started in calculus ii ,and I remember that most I didn't use specific rules when dealing with derivatives and I sometimes managed to get away with it(especially in physics. now I have been playing with relations and I got unacceptable results so i will post what I did ( even if all of it is wrong)
let y=f(x)
y''=dy'/dx=dy'/dy dy/dx

How did you go from dy'/dx to (dy'/dx)*(dy/dx )

 

[madah12]

I didnt go from dy'/dx to (dy'/dx)*(dy/dx) i went from dy'/dx to dy'/dy dy/dx

 

[dextercioby]

dy'/dy makes no sense. y' and y are independent <variables> (functions of x).

 

[madah12]

umm what about dv/dx in physics it would be meaningful to say the rate of the change of velocity with respect to the displacement is ... right?

 

[dextercioby]

No, no, you're using redundant notation and the integration is wrong

Let's use the notations in physics, because they have different letters.

a= \frac{1}{2}\frac{dv^2}{dx} but this happens only through v(t) => v(t(x)). Then

\frac{a}{v} = \frac{dv}{dx} then

\frac{1}{v}\frac{dv}{dt} = \frac{dv}{dt} \frac{1}{\frac{dx}{dt}} = \frac{1}{v}\frac{dv}{dt}

redundancy.

OK, let's choose your expression

\frac{1}{v} \frac{dv}{dt} = \frac{dv}{dx}. Multiply by the 1-form dt and integrate on a path in the v-line (LHS) and in the t-line (RHS)

\ln v(t) + C = \int\left[\left(\frac{dv(x)}{dx}\right)(t)\right] \, dt

There's no 1 in the RHS. First you compute the derivative of v wrt x and consider the result as a function of t.

 

[madah12]

in the last line in the right hand side you wrote
integral dv(x)/dx (t)dt

what does the (t) after /dx indicate?

 

[dextercioby]

The dependence of t of the function after differentiation. I edited the formula to make it clearer.

 

[SammyS]

...
1/y' dy' = dy'/dy *dx
1/y' dy' = d/dy * dy/dx *dx
and I canceled the dy and the dx
and got
ln(y')=1
I know this is stupid but how many operations I did were wrong?
...

dy'/dy This is the derivative of y' WRT y, where y'=dy/dx In the second line you write it as:
d/dy * dy/dx . However, that's not a multiplication. d/dy is an "operator", so you can't cancel dy. This is the rate of change in y' WRT y.

 

[madah12]

dy'/dy This is the derivative of y' WRT y, where y'=dy/dx In the second line you write it as:
d/dy * dy/dx . However, that's not a multiplication. d/dy is an "operator", so you can't cancel dy. This is the rate of change in y' WRT y.

how do we know when it is a multiplication and when is it an operation I mean when you have
av=vdv/dt
<=> adx/dt = vdv/dt
you can cancel it out to get adx=vdv
which gives you vf^2 - vi^2=2ax

 

[dextercioby]

1. The last integration follows iff the a is a constant.
2. The <dt> doesn't cancel out, because it's not a denominator of a fraction. Instead one has

a \frac{dx}{dt} = v \frac{dv}{dt} \Leftrightarrow a x&#039;(t) = v v&#039;(t)

which is an equality between 0-forms. Multiply by the 1 form dt in both terms and then

a x&#039;(t) dt = v v&#039;(t) dt.

Now use the definition of an exterior differential acting on a 0-form and have

a dx = v dv.

If a is constant (or a known function of <x>), then the integration can be perfomed in both sides of the last equation.

 

[madah12]

what is the exterior differential acting on a 0-form ? and what is a 0 form

 

[dextercioby]

That's part of the geometric interpretation of calculus. It's learned in a differential geometry class. For simplicity, you can do without though and assume dy(x) to be the differential of the function y(x) and is defined through

dy(x) = y&#039;(x) dx

where dx is the differential of the variable x.

 

[HallsofIvy]

Homework Statement

I just started in calculus ii ,and I remember that most I didn't use specific rules when dealing with derivatives and I sometimes managed to get away with it(especially in physics. now I have been playing with relations and I got unacceptable results so i will post what I did ( even if all of it is wrong)
let y=f(x)
y''=dy'/dx=dy'/dy dy/dx
=y' dy'/dy
y''/y'=dy'/dy
1/y' *dy'/dx = dy'/dy
1/y' dy' = dy'/dy *dx
1/y' dy' = d/dy * dy/dx *dx

You have an error here. You have (1/y')dy= d/dy(dy/dx) dx
That is, the "dy/dx" is inside the "d/dy" not just "multiplied by it".
You cannot cancel the "dx"s here.

and I canceled the dy and the dx
and got
ln(y')=1
I know this is stupid but how many operations I did were wrong?

Homework Equations

The Attempt at a Solution