Hello, i'm looking for someone who both knows the theory

>>8197041
Derivatives are operators, like +,-,×,÷ etc...

There's formal proofs that show why you can move them around like terms, but it would be too tedious, and unnecessary to show that every single time you work a problem, so you just use the shorthand

>>8197041
>can move dx around i
dy and dy are like small numbers that shrink away when push comes to shove (in certain 'calm' situations)
The theory why this actually works is complicated (see 'Hyperreal number')

say y = x*x

when y changes by dy, x changes by dx
so we get

y + dy = (x + dx)(x + dx)
= x*x + 2*x*dx + dx*dx

when dx gets small, dx*dx is a second order smallness and can be discarded

y + dy = x*x + 2*x*dx

remember, y = x*x so we have

y + dy = y +2*x*dx

dy = 2*x*dx

dy/dx = 2x

>>8197132
>dy/dx = 2x
Heh, that's pretty good.

y = sinx
y + dy = sin(x + dx)
y + dy = (sinx)(cosdx) + (cosx)(sindx)
y + dy = ycosdx + (cosx)(sindx)

hmm...

Well, looks like that's as far as the theory goes. Back to real calculus.

>>8197165
Cosdx is 1 to first order. Sin dx is dx to first order.

>>8197189
y + dy = y + (cosx)dx
dy = dxcosx
dy/dx = cosx

ayy.

>>8197189
>Cosdx is 1 to first order.
>Sin dx is dx to first order.
Why though?

It looks like this was made after calculus and those ruels were simply made so that this process would give the same result as calculus.

>>8197194
Taylor series, limits?

>>8197195
How the fuck are you going to construct a Taylor series without first knowing what a derivative is?

>>8197041
The answer is that you can't really do it if all you know is the definition of a derivative. If [math]y[/math] is a function of [math]x[/math], we define the derivative as[eqn]\frac{dy}{dx}(x_0) := \lim_{x \to x_0} \frac{y(x) - y(x_0)}{x - x_0}[/eqn]and a priori [math]\frac{d}{dx}[/math] is all one big symbol and can't be broken up.

First you might look at things like the chain rule, [math]\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}[/math], and see that they look like you're just multiplying the top and bottom by a [math]du[/math], even though the proof is more involved than that.

Then you might learn about differential forms or nonstandard analysis or any of a multitude of other fields which have [math]d[/math]'s that act in a similar way, e.g. [math]dy = y'(x) \,dx[/math], and try to consider [math]\frac{dy}{dx}[/math] as the "coefficient" (here a function of [math]x[/math]) of [math]dx[/math] in [math]dy[/math].

The point is that you have to use more sophisticated machinery to say why you can "multiply both sides by [math]dx[/math]" because the original definition of derivative doesn't even give a meaning to that symbol. In some sense they are acting like infinitesimal changes in [math]x[/math] and [math]y[/math] so any manipulations that keep that viewpoint in mind will be valid, but you'll have to formalize that sentiment if you want to use it like that. (This is pretty much nonstandard analysis.)

>why derivative notation stuff can be used both as a way to i guess represent something, and be used algebraically to manipulate or rearrange equations to be more convenient
This is the point of math, really. Model things with notation, and then describe valid operations we can do to the notation, and then use and abuse those allowed rules until we get an answer.

>>8197165
>Back to real calculus.

Silvanus Thompson: Calculus Made Easy
pages 17-18

http://www.gutenberg.org/ebooks/33283

>>8197243
Sines & cosines: page 163

>>8197165
rookie mistake

>>8197232
Sorry to sperg out on you but you forgot one of the most crucial properties a function should have in order to be differentiable. That is, if the limit you wrote in LaTeX exists, then the function must also be continuous at [math]x_0[/math]

>>8197872
Differentiability at a point implies the funciton is continuous at that point. Show for yourself: given that limit exists, pick the appropriate bounds to demonstrate continuity. It should be pretty easy if you multiply both sides by the denominator.

>>8197243
Thanks for the book senpai

>>8197243
>>8197258

How can we say that sin(1/2 dθ) = 1/2 dθ? If we regard dθ as infinitely small, surely lim dθ -> 0 of sin(1/2 dθ) = 0?

>>8198133
The explanation in the book essentially assumes knowledge about the rate of change of sin and cos near 0. One nice way to think about it by looking at the graph of sin, and noting that the function is very close to the function f(x) = x near 0.

In fact, the taylor expansion of sin is as follows:
[math]\sin(\theta) = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \dotsc[/math].
If you take this as a definition of the sin function, then it becomes clear that the reasoning for infinitesimals with sin is sensible.

>>8198166

Ah that makes a lot of sense, thanks.

>>8198133
No
Lim of delta theta approaches zero
Not the differential

>>8197243
This book has the greatest preface ever. Seriously, is there anything worse than an asshole with an over-inflated ego in academia?