# The Quotient Rule

What is the derivative of \( (x^2+1)/(x^3-3x)\)? 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 |
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Compute the derivative of \( (x^2+1)/(x^3-3x)\). SOLUTION \[ {d\over dx}{x^2+1\over x^3-3x}={2x(x^3-3x)-(x^2+1)(3x^2-3)\over(x^3-3x)^2}= {-x^4-6x^2+3\over (x^3-3x)^2}. \] |

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 |
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Find the derivative of \( \sqrt{625-x^2}/\sqrt{x}\) 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 \( \sqrt{x}=x^{1/2}\) to compute the derivative of \( \sqrt{x}\) 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 \[ -{x^2+625\over 2\sqrt{625-x^2}x^{3/2}}.\] |

{C}

Occasionally you will need to compute the derivative of a quotient with a constant numerator, like \( 10/x^2\). 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, \( x^2\) 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** \( {x^3\over x^3-5x+10}\) (answer)

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

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

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

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

**Ex 3.4.6** Find an equation for the tangent line to \( f(x) = (x-2)/(x^3 + 4x - 1)\) 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 \( y=1/(1+x^2)\) 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://instructional1.calstatela.edu/sgray/Agnesi/WitchHistory/Historynamewitch.html__. (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)