What is the derivative of $\ds (x^2+1)/(x^3-3x)$? More generally, we'd like to have a formula to compute the derivative of $f(x)/g(x)$ if we already know $f'(x)$ and $g'(x)$. Instead of attacking this problem head-on, let's notice that we've already done part of the problem: $f(x)/g(x)= f(x)\cdot(1/g(x))$, that is, this is "really'' a product, and we can compute the derivative if we know $f'(x)$ and $(1/g(x))'$. So really the only new bit of information we need is $(1/g(x))'$ in terms of $g'(x)$. As with the product rule, let's set this up and see how far we can get: $$ \eqalign{ {d\over dx}{1\over g(x)}&=\lim_{\Delta x\to0} {{1\over g(x+\Delta x)}-{1\over g(x)}\over\Delta x}\cr &=\lim_{\Delta x\to0} {{g(x)-g(x+\Delta x)\over g(x+\Delta x)g(x)}\over\Delta x}\cr &=\lim_{\Delta x\to0} {g(x)-g(x+\Delta x)\over g(x+\Delta x)g(x)\Delta x}\cr &=\lim_{\Delta x\to0} -{g(x+\Delta x)-g(x)\over \Delta x} {1\over g(x+\Delta x)g(x)}\cr &=-{g'(x)\over g(x)^2}\cr }$$ Now we can put this together with the product rule: $${d\over dx}{f(x)\over g(x)}=f(x){-g'(x)\over g(x)^2}+f'(x){1\over g(x)}={-f(x)g'(x)+f'(x)g(x)\over g(x)^2}= {f'(x)g(x)-f(x)g'(x)\over g(x)^2}. $$

Example 3.4.1 Compute the derivative of $\ds (x^2+1)/(x^3-3x)$. $${d\over dx}{x^2+1\over x^3-3x}={2x(x^3-3x)-(x^2+1)(3x^2-3)\over(x^3-3x)^2}= {-x^4-6x^2+3\over (x^3-3x)^2}. $$

$\square$

It is often possible to calculate derivatives in more than one way, as we have already seen. Since every quotient can be written as a product, it is always possible to use the product rule to compute the derivative, though it is not always simpler.

Example 3.4.2 Find the derivative of $\ds \sqrt{625-x^2}/\sqrt{x}$ in two ways: using the quotient rule, and using the product rule.

Quotient rule: $${d\over dx}{\sqrt{625-x^2}\over\sqrt{x}} = {\sqrt{x}(-x/\sqrt{625-x^2})-\sqrt{625-x^2}\cdot 1/(2\sqrt{x})\over x}.$$ Note that we have used $\ds \sqrt{x}=x^{1/2}$ to compute the derivative of $\ds \sqrt{x}$ by the power rule.

Product rule: $${d\over dx}\sqrt{625-x^2} x^{-1/2} = \sqrt{625-x^2} {-1\over 2}x^{-3/2}+{-x\over \sqrt{625-x^2}}x^{-1/2}. $$

With a bit of algebra, both of these simplify to $$-{x^2+625\over 2\sqrt{625-x^2}x^{3/2}}.$$

$\square$

Occasionally you will need to compute the derivative of a quotient with a constant numerator, like $\ds 10/x^2$. Of course you can use the quotient rule, but it is usually not the easiest method. If we do use it here, we get $${d\over dx}{10\over x^2}={x^2\cdot 0-10\cdot 2x\over x^4}= {-20\over x^3},$$ since the derivative of 10 is 0. But it is simpler to do this: $${d\over dx}{10\over x^2}={d\over dx}10x^{-2}=-20x^{-3}.$$ Admittedly, $\ds x^2$ is a particularly simple denominator, but we will see that a similar calculation is usually possible. Another approach is to remember that $${d\over dx}{1\over g(x)}={-g'(x)\over g(x)^2},$$ but this requires extra memorization. Using this formula, $${d\over dx}{10\over x^2}=10{-2x\over x^4}.$$ Note that we first use linearity of the derivative to pull the 10 out in front.

## Exercises 3.4

Find the derivatives of the functions in 1–4 using the quotient rule.

**Ex 3.4.1**
$\ds {x^3\over x^3-5x+10}$
(answer)

**Ex 3.4.2**
$\ds {x^2+5x-3\over x^5-6x^3+3x^2-7x+1}$
(answer)

**Ex 3.4.3**
$\ds {\sqrt{x}\over\sqrt{625-x^2}}$
(answer)

**Ex 3.4.4**
$\ds {\sqrt{625-x^2}\over x^{20}}$
(answer)

**Ex 3.4.5**
Find an equation for the tangent line to $\ds f(x) = (x^2 -
4)/(5-x)$ at $x= 3$.
(answer)

**Ex 3.4.6**
Find an equation for the tangent line to
$\ds f(x) = (x-2)/(x^3 + 4x - 1)$ at $x=1$.
(answer)

**Ex 3.4.7**
Let $P$ be a polynomial of degree $n$ and let $Q$ be a
polynomial of degree $m$ (with $Q$ not the zero polynomial).
Using sigma notation we can write
$$P=\sum _{k=0 } ^n a_k x^k,\qquad
Q=\sum_{k=0}^m b_k x^k.
$$
Use sigma notation to write the derivative of the
**rational function**
$P/Q$.

**Ex 3.4.8**
The curve $\ds y=1/(1+x^2)$ is an example of a class of
curves each of which is called a **witch of
Agnesi**.
Sketch the curve and find the tangent line to the curve at
$x= 5$. (The word *witch* here is a mistranslation of the
original Italian, as described at
http://mathworld.wolfram.com/WitchofAgnesi.html and
http://witchofagnesi.org/.
(answer)

**Ex 3.4.9**
If $f'(4) = 5$, $g'(4) = 12$, $(fg)(4)= f(4)g(4)=2$, and $g(4) = 6$,
compute $f(4)$ and $\ds{d\over dx}{f\over g}$ at 4.
(answer)