Check out this dialogue between two calculus students (based on a true story):

Devyn

I hate the limit definition of derivative. I wish there were a shorter way.

Riley

I think I might have found a pattern for taking derivatives.

Devyn

Really? I love patterns!

Riley

I know! Check this out, I’ve made a chart

\[ \begin{array}{c|c} f(x) & f'(x)\\ \hline x^2 & 2\cdot x^1\\ x^3 & 3\cdot x^2\\ x^4 & 4\cdot x^3 \end{array} \]

So maybe if we have a function

\[ f(x) = x^n\quad \text {then}\qquad f'(x) = n\cdot x^{n-1}. \]

Devyn

Hmmm does it work with square roots?

Riley

Oh that’s right, a square root is a power, just write

\[ f(x) = \sqrt {x} = x^{1/2}. \]

So a square root is of the form \(x^n\).

Devyn

Let’s check it. If \(f(x) = \sqrt {x}\), \begin{align*} f'(x) &= \lim _{h\to 0} \frac {\sqrt {x+h} -\sqrt {x}}{h}\\ &= \lim _{h\to 0} \left (\frac {\sqrt {x+h} - \sqrt {x}}{h}\cdot \frac {\sqrt {x+h} + \sqrt {x}}{\sqrt {x+h} + \sqrt {x}}\right )\\ &= \lim _{h\to 0} \frac {x+h - x}{h(\sqrt {x+h} + \sqrt {x})}\\ &= \lim _{h\to 0} \frac {\cancel {h}}{\cancel {h}(\sqrt {x+h} + \sqrt {x})}\\ &= \lim _{h\to 0} \frac {1}{\sqrt {x+h} + \sqrt {x}}\\ &= \frac {1}{\sqrt {x} + \sqrt {x}}\\ &= \frac {1}{2\sqrt {x}}\\ &= \frac {1}{2}\cdot x^{-1/2}. \end{align*}

Riley

Holy Cat Fur! It works! In this case \(f'(x) = n\cdot x^{n-1}\).

Devyn

I wonder if it always works? If so I want to know why it works! I wonder what other patterns we can find?

The pattern

holds whenever \(n\) is a constant. Explaining why it works in generality will take some time. For now, let’s see if we can use the problem to squash some derivatives with ease.