$f(z) = \frac{4 z + 4}{z^2 + z + 1}$

I have a question about the reciprocal rule concerning derivatives. I want to work around doing the quotient rule for the above function if possible, but I am not sure if a linear expression like $(4z + 4)$ can be taken out for $[1/(1z^2 +1z + 1)]$ to be used as "$v$" in the reciprocal rule $( (1/v)' = -(v'/v^2) )$.

Essentially my question is, can the numerator only contain a constant that is not 1, like 4 or pi, for said numerator to be "taken out" so the reciprocal rule can be applied? If not, why not and is there still a better way to find the derivative without doing a burdensome quotient rule here? Maybe the synthetic/long division method? Thanks for answers in advance.

  • 3
    $\begingroup$ You can always write $\frac{f(x)}{g(x)}$ as $f(x)\cdot \frac{1}{g(x)}$ and use Product Rule and Reciprocal Rule. Roughly as cumbersome as Quotient Rule, often worse, particularly when you want to know where the derivative is positive, negative, $0$. I suggest using the Quotient Rule $\endgroup$ – André Nicolas Jun 9 '14 at 23:57

No, you are not correct. Note that the quotient rule, given by the following for $u$ and $v$ functions of $x$ such that $v \neq 0$ $$\frac{\text{d}}{\text{d}x}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \,\,,$$ which reduces to the reciprocal rule for $u = 1$: $$\frac{\text{d}}{\text{d}x}\left(\frac{1}{v}\right) = -\frac{v'}{v^2} \,\,.$$ Thus, in this case, we cannot use the reciprocal rule, but we can apply the quotient rule just as easily to get $$\frac{\text{d}}{\text{d}z}f(z) = \frac{\text{d}}{\text{d}z}\left(\frac{4z+4}{z^2 + z + 1}\right) = \frac{4(z^2 + z + 1) - (4z + 4)(2z + 1)}{(z^2 + z + 1)^2} = -\frac{4z(z+2)}{(z^2+z+1)^2}\,.$$ In general, you can apply the reciprocal rule for differentiation of a fraction when, in the setup given above, $u = C$ for some constant $C$, since $\frac{\text{d}}{\text{d}x} C = 0$, and additionally $v$ is nonzero (as a special case of the quotient rule). This can be seen as follows $$\frac{\text{d}}{\text{d}x}\left(\frac{C}{v}\right) = C\frac{\text{d}}{\text{d}x}\left(\frac{1}{v}\right) = -C\frac{v'}{v^2}\,\,,$$ which can be viewed as a constant multiple of the result of an application of the reciprocal rule.


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