Why Quotients Need Care
A quotient can combine familiar functions while having restrictions that neither function has on its own. For instance, \(e^x\) is defined for every real \(x\), but \(\ln x\) is defined only for \(x>0\), and a quotient with \(\ln x\) in its denominator also excludes the input where \(\ln x=0\).
To differentiate such expressions reliably, first identify the domain, then apply the quotient rule, and finally make sure the derivative is stated only where the original function is defined. These checks matter because a derivative formula cannot restore inputs that the original quotient excludes.
The Quotient Rule
The condition \(v(x)\ne 0\) is essential: the quotient itself is not defined where its denominator is zero. In applications, also check that both \(u\) and \(v\) are defined and differentiable at the input.
Why the Quotient Rule Works
The rule follows from the limit definition of a derivative. Write \(q(x)=u(x)/v(x)\), and suppose \(v(x)\ne0\). For sufficiently small \(h\), differentiability implies that \(v(x+h)\) is close to \(v(x)\), so \(v(x+h)\ne0\) as well. Then the difference quotient can be combined over a common denominator:
Separate the two difference quotients and take the limit as \(h\to0\). Differentiability gives the limits \(u'(x)\) and \(v'(x)\), while \(v(x+h)\to v(x)\). Therefore,
This derivation explains both the order of the terms in the numerator and the squared denominator. Reversing the subtraction changes the sign and gives an incorrect derivative.
Domain Checks for Exponential and Logarithmic Quotients
The natural exponential \(e^x\) is defined and positive for every real \(x\). The natural logarithm \(\ln x\) is defined only for \(x>0\). In addition, any expression used as a denominator must be nonzero.
| Quotient | Domain restriction | Reason |
|---|---|---|
| \(\dfrac{e^x}{\ln x}\) | \(x>0\) and \(x\ne1\) | \(\ln x\) requires \(x>0\), and \(\ln 1=0\). |
| \(\dfrac{\ln x}{a^x}\), \(a>0\) | \(x>0\) | \(\ln x\) requires \(x>0\); \(a^x\) is positive and never zero. |
| \(\dfrac{a^x}{\ln x}\), \(a>0\) | \(x>0\) and \(x\ne1\) | \(\ln x\) is the denominator and vanishes at \(x=1\). |
For a positive base \(a\), the derivative formula used below is \(\frac{d}{dx}a^x=a^x\ln a\). The natural logarithm has derivative \(\frac{d}{dx}\ln x=1/x\), for \(x>0\). These are the basic derivatives needed for the examples.
Worked Example: The Natural Exponential over the Natural Logarithm
Worked Example: The Natural Exponential over the Natural Logarithm
Differentiate \(f(x)=\dfrac{e^x}{\ln x}\), and evaluate \(f'(e)\).
Step 1: Find the domain. The logarithm requires \(x>0\). The denominator must also be nonzero, so \(\ln x\ne0\), which excludes \(x=1\). Thus the domain is \(x>0,\ x\ne1\).
Step 2: Identify the numerator, denominator, and derivatives. Let \(u(x)=e^x\) and \(v(x)=\ln x\). Then \(u'(x)=e^x\) and \(v'(x)=1/x\).
Step 3: Apply the quotient rule.
This derivative is defined on the same domain, \(x>0,\ x\ne1\).
Step 4: Evaluate at \(x=e\). Since \(\ln e=1\), the quotient-rule numerator at \(e\) is \(1\cdot e^e-e^e\cdot(1/e)=e^e-e^{e-1}\), and the squared denominator is \(1^2=1\). Therefore,
Check: Substituting \(x=e\) into the unsimplified quotient-rule expression gives \(\frac{e^e\ln e-e^e/e}{(\ln e)^2}=\frac{e^e-e^{e-1}}{1}\), agreeing with the result.
Worked Example: The Natural Logarithm over an Exponential
Worked Example: The Natural Logarithm over an Exponential
Differentiate \(g(x)=\dfrac{\ln x}{2^x}\), and find \(g'(1)\).
Step 1: Find the domain. We need \(x>0\) for \(\ln x\). Since \(2^x>0\) for every real \(x\), it never makes the denominator zero. The domain is \(x>0\).
Step 2: Identify the derivatives. Let \(u(x)=\ln x\) and \(v(x)=2^x\). Then \(u'(x)=1/x\) and \(v'(x)=2^x\ln2\).
Step 3: Use the quotient rule and simplify.
Step 4: Evaluate at \(x=1\). We have \(1/x=1\), \(\ln1=0\), and \(2^1=2\). Thus
Check: In the quotient-rule form, the numerator at \(1\) is \(2(1)-0\cdot(2\ln2)=2\), and the denominator is \(2^2=4\). The result is \(2/4=1/2\), as above.
Worked Example: Another Base with a Logarithm in the Denominator
Worked Example: Another Base with a Logarithm in the Denominator
Differentiate \(h(x)=\dfrac{3^x}{\ln x}\), and evaluate the derivative at \(x=e\).
Step 1: Find the domain. The logarithm requires \(x>0\), and its value cannot be zero in the denominator. Since \(\ln x=0\) at \(x=1\), the domain is \(x>0,\ x\ne1\).
Step 2: Differentiate the numerator and denominator. Set \(u(x)=3^x\) and \(v(x)=\ln x\). Then \(u'(x)=3^x\ln3\) and \(v'(x)=1/x\).
Step 3: Apply the rule.
Step 4: Evaluate at \(e\). Using \(\ln e=1\) and \(1/e\) for the reciprocal of \(e\),
Check: At \(e\), the top derivative is \(3^e\ln3\), the bottom value is \(1\), the top value is \(3^e\), and the bottom derivative is \(1/e\). The quotient-rule numerator is therefore \(1(3^e\ln3)-3^e(1/e)\), matching the expression found.
A Useful Simplification—and Its Limit
In the example \(g(x)=\ln x/2^x\), the quotient-rule result simplified because one factor of \(2^x\) canceled against the squared denominator. More generally, for \(a>0\),
This simplification is valid because \(a^x\ne0\). In contrast, canceling or simplifying an expression involving \(\ln x\) in the denominator does not permit \(x=1\): the original quotient \(e^x/\ln x\) is still undefined there. A simplified derivative formula must always retain the original function's domain restrictions.
Common Mistake: Reversing the Numerator
A frequent error is to reverse the subtraction in the quotient rule. For \(f(x)=e^x/\ln x\), the incorrect calculation might be
This is the negative of the correct numerator. The rule is denominator times derivative of numerator minus numerator times derivative of denominator. With \(u=e^x\) and \(v=\ln x\), that order gives
Another domain error is to report this derivative at \(x=1\). Although \(e^x\) is defined there, \(\ln1=0\), so the original quotient is undefined. Consequently, it has no derivative at \(x=1\).
A Reliable Process
Check logarithm inputs and exclude every input that makes a denominator zero.
Write \(u\) for the numerator and \(v\) for the denominator, and find \(u'\) and \(v'\).
Form \(vu'-uv'\), then divide by \(v^2\).
Cancel only valid nonzero factors, and keep the derivative restricted to the original domain.
These quotients are useful practice because they require both accurate differentiation and careful attention to where the expressions exist. The most important checks are simple: the logarithm's input must be positive, the denominator must not vanish, and the quotient-rule subtraction must remain in the correct order.
Check Your Understanding
For each question, state any relevant domain restriction before differentiating or evaluating.
- Find the domain of \(p(x)=e^x/\ln x\). Why is \(x=1\) excluded?
- Differentiate \(r(x)=\ln x/5^x\) and simplify the result.
- Differentiate \(s(x)=4^x/\ln x\), and state its domain.
- Why is \(2^x\) never a zero denominator, while \(\ln x\) is zero at \(x=1\)?
- For \(t(x)=\ln x/3^x\), calculate \(t'(1)\) using the quotient rule.
