Patterns in derivatives

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Two young mathematicians think about “short cuts” for differentiation.

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 $f(x) = \sqrt {x}$ , $\begin{align*} f'(x) &= \lim_{h\to 0} \frac{\sqrt{x+h} -\sqrt{x}}{h}\\ &= \lim_{h\to 0} \left(\frac{\sqrt{x+h} - \sqrt{x}}{h}\cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}\right)\\ &= \lim_{h\to 0} \frac{x+h - x}{h(\sqrt{x+h} + \sqrt{x})}\\ &= \lim_{h\to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}\\ &= \lim_{h\to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}\\ &= \frac{1}{\sqrt{x} + \sqrt{x}}\\ &= \frac{1}{2\sqrt{x}}\\ &= \frac{1}{2}\cdot x^{-1/2}. \end{align*}$
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.

Using the pattern found above, compute: $\ddx x^{101} \begin {prompt}= \answer {101 x^{100}}\end {prompt}$

Using the pattern found above, compute: $\ddx \frac {1}{x^{77}} \begin {prompt}= \answer {-77 x^{-78}}\end {prompt}$

Using the pattern found above, compute: $\ddx \sqrt [5]{x} \begin {prompt}= \answer {x^{-4/5}/5}\end {prompt}$

Using the pattern found above, compute: $\ddx x^e \begin {prompt}= \answer {e x^{e-1}}\end {prompt}$