Two Similar Expressions with Different Meanings
The expressions \(f'(x)/g'(x)\) and \((f(x)/g(x))'\) look related: both involve derivatives and a quotient. But they are generally not equal. This distinction matters whenever you use L'Hôpital's Rule, because that rule compares the derivatives of the numerator and denominator; it does not ask you to differentiate the quotient \(f/g\).
For clarity, \(f'(x)/g'(x)\) means “differentiate \(f\), differentiate \(g\), then divide.” By contrast, \((f(x)/g(x))'\) means “first form the quotient, then differentiate it.” The quotient rule shows exactly how the latter expression differs.
The Quotient Rule
If \(f\) and \(g\) are differentiable at \(x\), and \(g(x)\neq 0\), then the quotient \(f/g\) is differentiable there and
This formula contains both \(f'\) and \(g'\), but it is not their ratio. Its numerator is the difference \(f'g-fg'\), and its denominator is \(g^2\). The quotient \(f'/g'\), in contrast, has \(g'\) as its denominator and does not contain the factor \(f\) or the factor \(g\) in the numerator.
Why the Quotient Rule Has This Form
The formula follows by writing the quotient as a product and differentiating. At points where \(g(x)\neq 0\), we can write \(f/g=f g^{-1}\). The product rule and chain rule then give
The condition \(g(x)\neq 0\) is essential: both the original quotient and the reciprocal \(g^{-1}\) require a nonzero denominator. The displayed formula describes the derivative of the quotient wherever the quotient is defined. It does not become the derivative ratio \(f'/g'\) merely because both formulas involve derivatives.
Why L'Hôpital's Rule Uses a Derivative Ratio
L'Hôpital's Rule gives a way to evaluate certain limits of \(f(x)/g(x)\) by studying \(f'(x)/g'(x)\). For the finite-endpoint \(0/0\) case, one standard form is as follows: suppose \(f\) and \(g\) are continuous at \(a\), differentiable on a one-sided interval next to \(a\), and \(f(a)=g(a)=0\). Suppose also that \(g(x)\neq 0\) and \(g'(x)\neq 0\) on that punctured interval. If
exists, with \(L\) finite, then
The relevant comparison is between the changes in \(f\) and \(g\) near \(a\). The Cauchy Mean Value Theorem explains this relationship. Fix \(x\) on the interval near \(a\). Under the stated continuity and differentiability conditions, there is a point \(c\) between \(a\) and \(x\) such that
Since \(f(a)=g(a)=0\), and \(g(x)\neq 0\) and \(g'(c)\neq 0\), this equality can be divided by \(g(x)g'(c)\). It yields
As \(x\) approaches \(a\), the point \(c\), which lies between \(a\) and \(x\), also approaches \(a\). Thus, if \(f'/g'\) tends to \(L\) near \(a\), the right-hand side tends to \(L\), and so does \(f/g\). This argument explains why the derivative ratio appears: it compares the functions' changes through the Mean Value Theorem. It does not differentiate their quotient.
Worked Examples: Keep the Operations Separate
Worked Example: A Removable Quotient with Different Derivatives
Compare the derivative ratio and the derivative of the quotient for \(f(x)=x^2-1\) and \(g(x)=x-1\), near \(x=1\). First, \(f(1)=0\) and \(g(1)=0\), so the original quotient has the \(0/0\) form at that point.
Differentiate the numerator and denominator separately:
Therefore,
L'Hôpital's Rule gives the same limit for the original quotient. Directly, for \(x\neq 1\),
so \(\lim_{x\to 1}f(x)/g(x)=2\). Now differentiate the quotient itself. Using the quotient rule gives
for \(x\neq 1\). This also checks by differentiating the simplified quotient \(x+1\), whose derivative is \(1\). Thus the derivative ratio tends to \(2\), while the derivative of the quotient is \(1\).
Worked Example: The Derivative of the Quotient Has Its Own Limit
Let \(f(x)=e^x-1\) and \(g(x)=x\), and compare the two expressions as \(x\to 0\). Both functions vanish at zero. Differentiating them separately gives
L'Hôpital's Rule therefore gives \(\lim_{x\to 0}(e^x-1)/x=1\). But differentiating the quotient, for \(x\neq 0\), gives
To find the limit of this derivative, the numerator and denominator both tend to zero. Applying L'Hôpital's Rule to this new limit gives
The derivative ratio tends to \(1\), while the derivative of the quotient tends to \(1/2\). As a check on the quotient-rule calculation, differentiating its numerator \(xe^x-e^x+1\) gives \(e^x+xe^x-e^x=xe^x\), and differentiating its denominator \(x^2\) gives \(2x\), as used in the second application.
Worked Example: A Simple Error in a \(0/0\) Limit
Take \(f(x)=x^2\) and \(g(x)=x\), with \(x\to 0\). Since \(f(0)=g(0)=0\), the original quotient has the \(0/0\) form. The derivative ratio is
Indeed, the original quotient is \(x^2/x=x\) for \(x\neq 0\), so its limit is \(0\). The quotient-rule derivative, however, is
for \(x\neq 0\). This agrees with differentiating the simplified quotient \(x\), and confirms that the derivative of the quotient is not the derivative ratio.
A Common Mistake and Its Correction
A tempting but incorrect calculation is to write
For \(f(x)=x^2-1\) and \(g(x)=x-1\), this mistake says the derivative of the quotient is \(2x\). But the quotient rule gives \(1\) for \(x\neq 1\), as the first example verified. The error changes the operation: it replaces differentiating \(f/g\) with separately differentiating \(f\) and \(g\), then dividing.
A Practical Check Before You Calculate
Before using either expression, identify the operation in the question. If the task is to differentiate a quotient, write the quotient-rule formula and check that its denominator is nonzero. If the task is to evaluate a limit with L'Hôpital's Rule, first verify that the original limit has an applicable indeterminate form, then form \(f'/g'\) and check the rule's hypotheses. Do not insert the quotient-rule derivative into that comparison.
The two expressions can sometimes have the same value, but that does not make them interchangeable. For instance, coincidences may occur at particular points or for special choices of \(f\) and \(g\). The correct approach is to use the formula that matches the requested operation, rather than infer equality from a special case.
Check Your Understanding
For each question, keep the derivative ratio distinct from the derivative of a quotient.
- State the quotient-rule formula for \((f/g)'\), including the condition needed for the quotient to be defined.
- For \(f(x)=x^2\) and \(g(x)=x\), what are \(f'(x)/g'(x)\) and \((f(x)/g(x))'\) for \(x\neq 0\)?
- In the finite-endpoint \(0/0\) case, what limit does L'Hôpital's Rule compare to evaluate \(\lim f(x)/g(x)\)?
- Why does the Cauchy Mean Value Theorem lead to a ratio of derivatives rather than the derivative of \(f/g\)?
- Give one concrete error that results from writing \((f/g)'=f'/g'\), and state the correction.
