This section is aimed at students in upper secondary education in the Danish
school system, some objects will be simplified some details will be omitted.
Power Functions
Power functions are similar to exponential
functions in structure, but
rather different in nature. The only difference in structure is that you
"flip" the power, so that the variable becomes the base and the constant
becomes the exponent, as such
$$f(x)=bx^a$$
The most immediate difference in nature is the interpretation of the
constants, for rational \(a=\frac{n}{m}\) we have the interpretation that
\(x^{a}=\sqrt[m]{x^n}\) which has some domain issues if \(m\) is even and
\(x^n< 0\) since taking even roots of negative numbers yields complex
numbers as results. If we extend it to \(a\in\mathbb{R}\) it only becomes
more complicated, which together with the result that for any \(n< 0\),
\(f(0)=\infty\) will motivate us to restrict ourselves to \(x>0\) on
first approach, i.e. \(dom(f)=\mathbb{R}_+^*=\{x\in\mathbb{R}:x>0\).
For an interpretation of \(b\) consider that
$$f(1)=b⋅1^a=b⋅1=b$$
so it's not the initial value as such, but the value at 1, which feels much
less satisfactory than initial value for the exponential function, but such
is life.
Proposition(Nature of Growth)
As with other function types, the nature of the growth of power functions
is that if we increase the x-value by a specific percentage, then the
y-value will increase by another percentage dependent on \(a\), more
specifically do we have the following relation
$$r_y=(1+r_x)^a-1$$
To observe this consider that
$$f(x(1+r_x))=b(x(1+r_x))^a=bx^a(1+r_x)^a=f(x)(1+r_y)$$
This may come as no surprise, but power functions also have its own
two-point theorem.
Two-Point Theorem
Given two points \((x_1,y_1)\) and \((x_2,y_2\), there is a unique power
function who's graph passes through those two points and it has the
following constants
$$a=\frac{\log(y_2/y_1)}{\log(x_2/x_1)}=\frac{\log(y_2)-\log(y_1)}{\log(x_2)-\log(x_1)}$$
and
$$b=\frac{y_i}{x_i^a}$$
Proof
The proof starts out pretty identical to the others, by removing \(b\)
\begin{align}
&&\frac{y_2}{y_1}=&\frac{\cancel{b}x_2^a}{\cancel{b}x_1^a}=\left(
\frac{x_2}{x_1}\right)^a\\
\implies&&\log\left(\frac{y_2}{y_1}\right)=&\log\left(\left(
\frac{x_2}{x_1}\right)^a\right)=a\log\left(\frac{x_2}{x_1}\right)\\
\implies&&a=&\frac{\log(y_2/y_1)}{\log(x_2/x_1)}=\frac{\log(y_2)-
\log(y_1)}{\log(x_2)-\log(x_1)}
\end{align}
and reintroducing \(b\) we just find it by isolating it in the
function expression.
∎
Integer Power Derivative Theorem
Assume than we have a power function with non-zero integer exponent, i.e.
\(f(x)=x^n\), then its derivative is \(f'(x)=nx^{n-1}\). I've chosen
\(b=1\) since all it does is scale the derivative and can be added
afterwards.
Proof
We will per form a proof by induction, which entails confirming a
base case and the induction step. So lets take as our base case
\(n=1\), which yields \(f(x)=x\implies f'(x)=1=nx^{n-1}\) which
comforms to our statement. Now assume some integer
\(k\in\mathbb{N}\) also conforms to the statement and investigate
its successor \(n=k+1\). We start by writing the "rise" of the secant line
and then transitioning into the slope and then rewriting it into something
that depends on the secant slope for \(n=k\) and using our induction assumption.
\begin{align}
&&(x+h)^{k+1}-x^{k+1}=&(x+h)^k(x+h)-x^kx\\
&&=&(x+h)^kx+(x+h)^kh-x^kx\\
&&=&x((x+h)^k-x^k)+(x+h)^kh\\
\implies&&a=&x\frac{(x+h)^k-x^k}{h}+\frac{(x+h)^k\cancel{h}}{\cancel{h}}\\
&&\to&xkx^{k-1}+x^k\\
&&=&(k+1)x^k
\end{align}
if \(n< 0\) then we can perform a similar argument with \(n=-k\)
\begin{align}
&&(x+h)^{-k-1}-x^{-k-1}=&(x+h)^{-k}(x+h)^{-1}-x^{-k}x^{-1}\\
&&=&(x+h)^{-k}(x+h)^{-1}-x^{-k}(x+h)^{-1}+x^{-k}(x+h)^{-1}-x^{-k}x^{-1}\\
&&=&(x+h)^{-1}((x+h)^{-k}-x^{-k})+x^{-k}\left(\frac{1}{x+h}-\frac{1}{x}\right)\\
&&=&(x+h)^{-1}((x+h)^{-k}-x^{-k})+x^{-k}\left(\frac{\cancel{x}-(\cancel{x}+h)}{x(x+h)}\right)\\
\implies&&a=&(x+h)^{-1}\frac{((x+h)^{-k}-x^{-k}}{h}+x^{-k}\left(\frac{-\cancel{h}}{\cancel{h}x(x+h)}\right)\\
&&\to&x^{-1}(-kx^{k-1})-x^{-k}x^{-2}\\
&&=&(-k-1)x^{k-2}
\end{align}
∎
Real Powers
Now assume the power is \(a\in\mathbb{R}\), then we can write the function
as \(f(x)=x^a=e^{\ln(x^a)}=e^{a\ln(x)}\), which immediately yields the
following result by the formula for the derivative of a composite function
$$f'(x)=e^{a\ln(x)}a\ln'(x)=ax^a\frac{1}{x}=ax^{a-1}$$