The derivative of a power function

Differentiation, derivatives and Taylor approximations: Differentiating power functions

The derivative of a power function

First an example of a derivative of a power function, calculated via the basic definition of derivative.

Calculate the difference quotient of \(f(x)=x^{4}\) over the interval \([x,x+{\vartriangle}x]\) and determine the derivative function \(f'(x)\).

\[\begin{aligned}
\frac{{\vartriangle}f}{{\vartriangle}x} &= \frac{f(x+{\vartriangle}x)-f(x)}{{\vartriangle}x}\\ \\
&= \frac{(x+{\vartriangle}x)^{4}-x^{4}}{{\vartriangle}x} \\ \\
&= \frac{x^4+4x^3\cdot({\vartriangle}x)+6x^2\cdot({\vartriangle}x)^2+4x\cdot({\vartriangle}x)^3+({\vartriangle}x)^4-x^{4}}{{\vartriangle}x} \\ \\
&= 4x^3+6x^2\cdot({\vartriangle}x)+4x\cdot({\vartriangle}x)^2+({\vartriangle}x)^3\\ \\
&\to {4}x^{3}\quad\mathrm{if\;}{\vartriangle}x\to 0
\end{aligned}\]
So: \(f'(x)={4}x^{3}\)

New example

The pattern is clear after looking at enough examples.

Derivative of a power function The derivative of the function \(f(x)=x^p\) is equal to \(f'(x)=p\cdot x^{p-1}\) for any real number \(p\) and for any \(x\) in case the powers \(x^p\) and \(x^{p-1}\) have meaning.

If \(f(x)=\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}\), then \(f'(x)=-\tfrac{1}{2}x^{-\frac{1}{2}-1}=-\tfrac{1}{2}x^{-\frac{3}{2}}\) provided that \(x>0\).

For math enthusiasts we present two proofs.

In the first proof we assume that \(p\) is a positive integer and we use the following rules for differences of powers: \[a^p-b^p=(a-b)(a^{p-1}+a^{p-2}b+a^{p-3}b^2+\cdots+ab^{p-2}+b^{p-1})\] Then the derivative \(f'(a)\) at the point\(x=a\) for some \(a\) satisfies:
\[\begin{aligned}f'(a) &=\lim_{h\to 0}\frac{(a+h)^p-a^p}{h}\\[0.25cm]
&=\lim_{h\to 0}\frac{h\bigl((a+h)^{p-1}+(a+h)^{p-2}a+(a+h)^{p-3}a^2+\cdots+(a+h)a^{p-2}+a^{p-1}\bigr)}{h}\\[0.25cm]
&=\lim_{h\to 0} \bigl((a+h)^{p-1}+(a+h)^{p-2}a+(a+h)^{p-3}a^2+\cdots+(a+h)a^{p-2}+a^{p-1}\bigr)\\[0.25cm]
&=a^{p-1}+a^{p-2}a+a^{p-3}a^2+\cdots+a\cdot a^{p-2}+a^{p-1}\quad (p\text{ terms})\\[0.25cm]
&=p\cdot a^{p-1}\end{aligned}\]

In the second proof we anticipate on the chain rule for differentiation and the derivative of the natural logarithm, but the advantage is that that now \(p\) does not have to be a positive integer. If \[f(x)=x^p\] then \[\begin{aligned}\ln\bigl(f(x)\bigr)&= \ln(x^p)\\[0.25cm] &= p\cdot \ln(x)\end{aligned}\] From the chain rule for differentiation and the derivative of the natural logarithm it follows that \[\frac{f'(x)}{f(x)}=p\cdot \frac{1}{x}\] So: \[\begin{aligned}f'(x)&=p\cdot \frac{f(x)}{x}\\[0.25cm] &= p\cdot \frac{x^p}{x}\\[0.25cm] &=p\cdot x^{p-1}\end{aligned}\]

\(\phantom{x}\)

Mathcentre video

Differentiating Powers by First Principles (14:56)