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: $${d\over dx}{f(x)\over g(x)}=f(x){-g'(x)\over g(x)^2}+f'(x){1\over g(x)}={-f(x)g'(x)+f'(x)g(x)\over g(x)^2}= {f'(x)g(x)-f(x)g'(x)\over g(x)^2}. $$

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.

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 {\em witch\/} here is a mistranslation of the original Italian, as described at $$\hbox{\url{http://mathworld.wolfram.com/WitchofAgnesi.html} \vb|http://mathworld.wolfram.com/WitchofAgnesi.html|\endurl}$$ and $$\eqalign{ \hbox{\url{http://instructional1.calstatela.edu/sgray/Agnesi/WitchHistory/Historynamewitch.html} \vb|http://|\endurl} &\!\!\hbox{\url{http://instructional1.calstatela.edu/sgray/Agnesi/WitchHistory/Historynamewitch.html} \vb|instructional1.calstatela.edu/sgray/Agnesi/|\endurl}\cr &\hbox{\url{http://instructional1.calstatela.edu/sgray/Agnesi/WitchHistory/Historynamewitch.html} \vb|WitchHistory/Historynamewitch.html|\endurl.)}\cr }$$ (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)