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The Calculation of Slopes, Tangents, and Changes

Changes in smooth functions is what rate, or differential, equations are all about, be they trigonometric functions, exponential, or polynomial. As laws are introduced in text, they will relate some property of the matter with rates of change. The rate of change of a curve itself is a property of that curve, and will depend on the point of curve expressed as $x$, a value which is independent of the curve, and along with the values of the function in that neighborhood.

The slope of a function at a given point in the abscissa, $x$, is the derivative of that function, at $x$. The value of some function, $f$, for a given point on the abscissa, is near those values of the function, for some neighborhood $[x-b, x+b]$, if the function is smooth and continuous, on the entire interval.

Through a change of variable (COV) and a difference calculation we can formulate the derivative of any such continuous function. $x\to x+b$ indicates a substitution of $x+b$ for the original abscissa coordinate, $x$. The difference of this and the value of the function at $x$ is nominated, $\Delta f(x)$, and abbreviated as, $df$, and refers to the differential of the function, $f$. $$ d f(x) = f(x+b) - f(x) $$ That is the difference between $f$ at, or of, $x+b$ and $f$ at $x$, for some small quantity $b$, such that $b\ll (x_2-x_1)$, where we have chosen $x_2>x_1$, and could more safely state the absolute values of the element of change and the interval length are related by a much-less-than sign.

To get the quantity of physical significance (the derivative in variable of interest $x$, signified by $D_x$) we divide, or multiply by the inverse of, $b$, to get the rise-over-run at the offset-origin point $x$. So, that's a triangle with its hypotenuse as the tangent of interest, and the vertical right-side (there's three sides) as the $\Delta f$ (could be plus or minus), with the horizontal right-side extending from the point $(x, f(x))$ to the $(x+b, f(x))$ point which is the distance $b$. $$ D_x(f(x)) \equiv \frac{d}{dx} f(x)= \lim_{b\ll 1} \Delta f(x,b)/b = \lim_{b\ll1} (f(x+b)-f(x))b^{-1} $$ Which is the definition of the tangent of a function $f$, at $x$. The tangent is independent of the infinitesimal, $b$, because higher order terms, appearing neatly for polynomials, are many orders of magnitude smaller than the terms appearing with one factor of $b$. $b$ is defined in terms of the interval of the function, and is also referred to as $\Delta x$, or $\delta x$, the infinitesimal measure on $x$. So, we will calculate the tangent for $f(x)=x^n$, first for exponential power $n\gt-1$: $$ \Delta f(x,b) = (x+b)^n - x^n $$ Which appears as some terms of $b$ to an exponential power $q$ which itself is between $1$ and $n$ (the formulation of the terms of a cardinally-powered binomial being a subject of pre-17th century study, the Binomial theorem). The Binomial theorem is basically a formula for the terms of such a binomialbeing a summation kernel, or compactly $\sum\limits_{n=1}^m A_k x^{m-k}b^k$ where the $A_k$ is referred to as $m$-choose-$k$ cofactor.

The formula is derived as follows, using Leibniz' convention for $\Delta$ which is to mean the value which is the change in whatever is contiguously (not separated by $+$ or $+(-)=-$) to its right: given $f(x)=x^n$ then we calculate the change in $f$ at $x$ (any variable) for an arbitrary deviation from $x$ by $\Delta x$, the arbitrariness in this case will be such that the distance from $x$ will be negligable, by algebraic method. One can think about the derivative of $f_n$ at $x$ as being associated with a right-triangle with the tangent being the hypotenuse. which simultaneously adjusts both the vertical and horizontal sides $$ \Delta f(x) = f(x+\Delta x) - f(x) = (x + \Delta x)^n - x^n $$ Which says that if we want to know the difference in $f$ at $x$ and $f$ at some lesser/greater amount $\Delta x$ away from $x$, then we need a couple more tools. First we need the Binomial theorem, for cardinal number $n\geq 0$ (first published by Bhāskara II in 1150, along with the factorial concept): $(a + b)^n= n! \sum\limits_{k=0}^n (k!(n-k)!)^{-1} a^{n-k} b^k$ The factorial notation ($!$) was introduced in 1808 and is a shorthand for the number which is a multiplicative sequence $\prod\limits_1^n q$ (where $ab=ba$ is the commutative property of integers and reals, and is used ($(1)(2)\cdots(n)=(n)(n-1)\cdots(1)$)): $$ q!=(q)(q-1)(q-2)\cdots(q-q+1) $$ Where the last term can also be written $(q-(q-1))$, being useful for organization. For $q=5$, we have $5!=(5)(4)(3)(2)(1) = 120$. Also, the zero-factorial is defined as unity. Applying this yields a summation of $n$ terms (from the $n$ factors of the binomial $a+b$ or the binomial $x+\Delta x$), each being of the form $A_k x^n (\Delta x)^{n-k}$.

It may seem an intractable formula, but if one remembers that $\Delta x$ is the distance from our point of reference $x$, then it becomes clear that we can neaten up this formula if we designate the deviation as nominal only, so that $(\Delta x)^2 \ll (\Delta x)^1$ (i.e. the square of the infinitesimal is much much less than the single factor of the infinitesimal), which is true in so far as the infinitesimal is much much less than unity, $\Delta x \ll 1$. And so it is true that more numerous factors of $\Delta x$, are (more) negligable (than the squared infinitesimal term). So we only keep the first two terms of the binomial expansion, labelling with the big-o ($\mathcal{O}$) the largest of the terms we are omitting from consideration: $$ \Delta f= (n!)[ (0!n!)^{-1}x^n (\Delta x)^0 + (1!(n-1)!)^{-1}x^{n-1}(\Delta x)^1 + \mathcal{O}(\Delta x^2) ] -x^n = (n!(n!)^{-1}x^n + (n!)((n-1)!)^{-1}x^{n-1}(\Delta x)^1 + \mathcal{O}(\Delta x^2) ] -x^n $$ And since $q^0=1$, $(n!)(n!)^{-1}$ is just unity, and $(n!)((n-1)!)^{-1}=n$, we have: $$ \Delta f= x^n + nx^{n-1}\Delta x + (-x^n) = nx^{n-1} \Delta x $$ So in the limit $\Delta x\to0$, the derivative gives us the slope of the function precisely at the point $x$. Since the slope of the tangent $\equiv$ the rise-over-run, $\Delta f/ \Delta x$, for any term of the function which goes like $Ax^n$ in $x$ there is contributed a term $nAx^{n-1}$, since you can multiply the function in the derivation by $A$ without changing the dependence on $n$.

One can verify that the derivative of the anti-derivative is the same function of $x$, where the algebraic form of a vanishing term, $(q+1)(q+1)^{-1}=1$, is utilized. It isn't obvious that the slope of the Natural logarithm of $x$ is $x^{-1}$, but that the derivatives and anti-derivatives of every other power of $x$ is dictated by the above formula.


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