### A refresher on the quotient rule

The quotient rule is one of the more irritating and goofy things in elementary calculus, but it just couldn't have been any other way. The general principle is $${d\over dx}\left({f\over g}\right)={f'g-g'f\over g^2}$$ The main hazard is remembering that the numerator is as it is, rather than accidentally reversing the roles of $f$ and $g$, and then being off by $\pm$, which could be fatal in real life.

Here are some examples: \begin{align*} {d\over dx}\left({1\over x-2}\right) &= {{d\over dx}1\cdot(x-2)-1\cdot{d\over dx}(x-2) \over (x-2)^2}= {0\cdot (x-2)-1\cdot 1\over (x-2)^2}={-1\over (x-2)^2}\\ {d\over dx}\left({x-1\over x-2}\right)&= {(x-1)'(x-2)-(x-1)(x-2)'\over (x-2)^2}= {1\cdot(x-2)-(x-1)\cdot 1\over (x-2)^2}\\ &= {(x-2)-(x-1)\over (x-2)^2}= {-1\over (x-2)^2}\\ {d\over dx}\left({5x^3+x\over 2-x^7}\right)&= {(5x^3+x)'\cdot(2-x^7)-(5x^3+x)\cdot(2-x^7)\over (2-x^7)^2}\\ &= {(15x^2+1)\cdot(2-x^7)-(5x^3+x)\cdot(-7x^6)\over (2-x^7)^2} \end{align*} There's hardly any point in simplifying the last expression, unless someone gives you a good reason. In general, it's not so easy to see how much may or may not be gained in ‘simplifying’, and we won't make ourselves crazy over it.

#### Exercises

- Find ${ d \over dx }({ x-1 \over x-2 })$
- Find ${ d \over dx }({ 1 \over x-2 })$
- Find ${ d \over dx }({ \sqrt{x}-1 \over x^2-5 })$
- Find ${ d \over dx }({ 1-x^3 \over 2 + \sqrt{x} })$

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##### Calculus Refresher

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