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