I am very interested in math and I find calculus to be a particularly interesting subject, but one major problem I have with it is that I cannot find a consistent explanation of the rules of differentials (infantesimals) that explains all the things mathematicians do with them. I have truly looked everywhere, even in my expensive calculus books, but they just start using the notation and expect it to somehow make sense.(adsbygoogle = window.adsbygoogle || []).push({});

Therefore, I will try to explain my understanding of them and I would like you to point out where my errors are, since some of my understanding contradicts credible sites like wolfram:

First to develop some tools:

The differential is written

[tex]dx[/tex]

where x represents the variable that is to be varied by an infantesimaly small amount.

Another way to write a differential, usually when the variable or quantity to be varied is an expression is

[tex]d(x)[/tex]

where [tex]x[/tex] is the expression in question.

The operation of differentiation is NOT the same thing as finding the derivative (as it is defined on Wolfram and on Wikipedia) as a derivative is not a differential but a RATIO of differentials.

The derivative can be more accurately defined as first differentiation then dividing by the differential of the variable you want to find change with respect to.

For example, applying the operator [tex]\frac{d}{dx}[/tex] does not mean to differentiate, but rather to differentiate THEN divide by [tex]dx[/tex].

For example when given [tex]y = x[/tex] and asked to differentiate the correct answer is

[tex]dy = dx[/tex]

and when asked to find the derivative with respect to x this means first differentiate:

[tex]dy = dx[/tex]

then divide both sides by the differential of [tex]x[/tex]

[tex] \frac{dy}{dx} = \frac{dx}{dx}[/tex]

[tex]

\frac{dy}{dx} = 1

[/tex]

where the quantity [tex]\frac{dy}{dx}[/tex] is referred to as the 'derivative'.

The power rule is a special case of the chain rule and a distinction is made between them only because math professors are evil and wish to confuse students by making things seem more complicated:

[tex]y = (x)^n[/tex]

[tex]dy = n(x)^{n-1}dx[/tex]

If [tex]x[/tex] was a more complicated expression, as it is in the chain rule, we simply use direct substitution:

Say if [tex]x = 3u[/tex].

[tex]dy = n(3u)^{n-1}d(3u)[/tex]

[tex]dy = n(3u)^{n-1}(3)du[/tex]

Product rule:

[tex]y = f(x)g(x)[/tex]

[tex]dy = f'(x)(dx)g(x) + f(x)g'(x)(dx)[/tex]

Quotient rule:

[tex]y = \frac{f(x)}{g(x)}[/tex]

[tex]dy = \frac{g(x)f'(x)(dx) - f(x)g'(x)(dx)}{(g(x))^2}[/tex]

Differentiating a constant is zero:

[tex]d(c) = 0[/tex]

Differentiating a differential is described in notation as raising the d to higher and higher powers:

For example repeatedly differentiating y we get:

[tex]y[/tex]

[tex]dy[/tex]

[tex]d^{2}y[/tex]

[tex]d^{3}y[/tex]

The notation for a second derivative is

[tex]\frac{d^2y}{{dx}^2}[/tex]

This can be understood better as

[tex]\frac{d}{dx}\frac{dy}{dx}[/tex]

Here is a proof of this fact:

First differentiate dy/dx then divide by the quantity dx:

[tex]\frac{d}{dx}\frac{dy}{dx} = d{\frac{dy}{dx}}{\frac{1}{dx}}[/tex]

using the quotient rule:

[tex]d{\frac{dy}{dx}}[/tex] = low d high less high d low, underneath denominator squared will go =

[tex]\frac{(dx)(d^{2}y) - (dy)(d^{2}x)}{{dx}^2}[/tex]

Break this up into

[tex]\frac{d^2y}{dx} - \frac{(dy)(d^2x)}{{dx}^2}[/tex]

we can note that

[tex]\frac{(dy)(d^2x)}{(dx)^2} = 0[/tex]

because

[tex]\frac{(dy)(d^2x)}{(dx)^2} = (dy)(\frac{d}{dx})(\frac{dx}{dx})[/tex]

and since

[tex]\frac{dx}{dx} = 1[/tex]

[tex](dy)(\frac{d}{dx})(1) = 0[/tex]

since [tex]d(1) = 0[/tex]

Now that we know

[tex]d(\frac{dy}{dx}) = \frac{(d^2y)}{dx} - \frac{(dy)(d^2x)}{(dx)^2} = \frac{(d^2y)}{dx}[/tex]

we can say that

[tex](\frac{d}{dx})(\frac{dy}{dx}) = d(\frac{dy}{dx})(\frac{1}{dx}) = (\frac{d^2y}{dx})(\frac{1}{dx}) = \frac{d^2y}{{dx}^2}[/tex]

which is the standard notation for the second derivative.

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# Understanding differentials and differentiation

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