Why Domain and Derivative Signs Matter
Exponential and logarithmic functions often make derivative calculations relatively direct. Their behavior still depends on two checks that are easy to overlook: where the function is defined and how its derivative changes sign.
A logarithm requires a positive input, so a formula involving a logarithm may have a domain split into separate intervals. A point where the derivative formula is undefined is a critical number only if the original function is defined there. Once the domain is established, a derivative sign chart can show whether the function changes from decreasing to increasing, or from increasing to decreasing.
Derivative Rules and Domain Conditions
For a differentiable function \(u\), the chain rule gives the following formulas. In the logarithmic formulas, the input \(u(x)\) must be positive wherever the function is defined.
The exponential factors \(e^{u(x)}\) and \(a^{u(x)}\) are always positive. Also, \(\ln(a)\) is positive when \(a>1\) and negative when \(0<a<1\). These facts help determine derivative signs. In contrast, the denominator \(u(x)\) in a logarithmic derivative is positive on the logarithm’s domain, but the numerator \(u'(x)\) can change sign.
For instance, the derivative of \(\ln(u(x))\) has the same sign as \(u'(x)\) wherever \(u(x)>0\). For \(a^{u(x)}\), the sign also depends on \(\ln(a)\); if \(0<a<1\), that factor reverses the sign of \(u'(x)\).
In particular, a point outside the domain is not a critical number, even if a derivative expression has a zero denominator there. When the domain has separate intervals, derivative signs must be analyzed separately on those intervals. A sign change across a gap in the domain does not classify an extremum at a point in that gap.
How a Sign Change Classifies an Extremum
The First Derivative Test, established in the earlier material on derivative signs and local extrema, says that a change in the sign of \(f'\) classifies a critical point. A negative-to-positive change indicates a local minimum; a positive-to-negative change indicates a local maximum. Here is a short justification of the negative-to-positive case.
When applying this test, the intervals on each side of \(c\) must lie in the function’s domain. This condition matters especially for logarithmic expressions, whose domains can have gaps.
Worked Examples
Worked Example: An Exponential Function with a Local Minimum
Find and classify the critical numbers of \(f(x)=(x-1)e^x\).
The exponential is defined for every real \(x\), so the domain is \(\mathbb{R}\). Differentiate using the product rule:
Since \(e^x>0\) for every real \(x\), \(f'(x)=0\) exactly when \(x=0\). The derivative exists everywhere, so \(0\) is the only critical number. The sign of \(f'(x)=xe^x\) is the sign of \(x\): it is negative for \(x<0\) and positive for \(x>0\). Thus \(f\) changes from decreasing to increasing, and \(x=0\) is a local minimum.
The function value is \(f(0)=(0-1)e^0=-1\), so the local minimum value is \(-1\). Checking the critical-point calculation directly, \(f'(0)=0\cdot e^0=0\), and the sign change on the two sides confirms the classification.
Worked Example: A Logarithmic Function with a Local Maximum
Find and classify the critical numbers of \(f(x)=\frac{\ln x}{x}\).
The logarithm requires \(x>0\), so the domain is \((0,\infty)\). Using the quotient rule,
On the domain, \(x^2>0\). Therefore, the sign of \(f'(x)\) is the sign of \(1-\ln(x)\). Solving \(f'(x)=0\) gives \(\ln(x)=1\), hence \(x=e\). This point belongs to the domain, and \(f'\) exists there.
For \(0<x<e\), \(\ln(x)<1\), so \(f'(x)>0\). For \(x>e\), \(\ln(x)>1\), so \(f'(x)<0\). The function changes from increasing to decreasing, giving a local maximum at \(x=e\). Its value is
As a check, substituting \(x=e\) into the derivative gives \(f'(e)=(1-\ln(e))/e^2=0\). The signs on both sides establish that this critical point is a maximum, not a minimum.
Worked Example: A Logarithm with a Disconnected Domain
Find and classify the critical numbers of \(f(x)=\ln(x^2-1)-x\).
The logarithm’s input must be positive:
Thus the domain is \((-\infty,-1)\cup(1,\infty)\). Differentiating on this domain gives
The denominator is positive throughout the domain. Setting the numerator equal to zero gives \(x^2-2x-1=0\), so \(x=1-\sqrt{2}\) or \(x=1+\sqrt{2}\). The first root lies between \(-1\) and \(1\), where \(f\) is not defined. It is not a critical number. The second root, \(c=1+\sqrt{2}\), is in the domain and is the only critical number.
On \((-\infty,-1)\), the numerator \(-x^2+2x+1\) is negative, so \(f'(x)<0\). On \((1,c)\), that numerator is positive, while on \((c,\infty)\) it is negative. Therefore, the function increases before \(c\) within the right-hand domain interval and decreases after it. The point \(c\) is a local maximum. Its value is
To check the critical point, the numerator of \(f'(c)\) is \(-c^2+2c+1=0\), while \(c^2-1>0\); hence \(f'(c)=0\). The derivative remains negative throughout the separate left-hand interval, so it has no critical number there. The excluded points \(x=-1\) and \(x=1\) are not candidates because they are not in the domain.
Common Pitfall: Treating an Excluded Point as Critical
Consider \(g(x)=\ln(x-2)\). Its domain is \(x>2\), and its derivative is \(g'(x)=1/(x-2)\). A tempting but incorrect statement is: “The derivative is undefined at \(x=2\), so \(2\) is a critical number.” The definition requires a critical number to belong to the domain, and \(g(2)\) does not exist.
For every \(x>2\), the derivative \(1/(x-2)\) is positive. The function is increasing throughout its domain and has no critical number there. The correct procedure is to determine the domain first, then identify zeros and points of derivative failure that lie in that domain. A boundary excluded by a logarithm’s domain condition cannot be classified as a local extremum of the function.
A Reliable Approach
For exponential and logarithmic expressions, use the following sequence. It keeps domain restrictions and derivative signs visible at each stage.
For every logarithm, require its input to be positive. Record any separate intervals explicitly.
Apply the chain rule and retain factors such as \(e^{u(x)}\), \(a^{u(x)}\), and \(\ln(a)\) when determining signs.
Solve \(f'(x)=0\) and check that each solution belongs to the domain. Also check for points in the domain where the derivative does not exist.
Check the sign of the derivative on either side of each valid critical number without crossing a gap in the domain.
A negative-to-positive change gives a local minimum; a positive-to-negative change gives a local maximum. Substitute the critical number into the original function to find the extremum value.
Why These Checks Matter
In an exponential expression, a positive exponential factor often leaves the sign to be determined by a simpler factor. In a logarithmic expression, the derivative includes the derivative of the logarithm’s input, and that input must remain positive. These properties can make sign analysis efficient, but they do not remove the need to check the domain.
Finding local extrema also does not settle whether a function has a largest or smallest value over its entire domain. This tutorial’s sign analysis is local: it determines how the function behaves near its critical numbers. Comparing values across a whole interval or domain is a separate question.
Check Your Understanding
For each question, begin by stating the domain before using a derivative sign chart.
- For \(f(x)=(x+2)e^x\), find the critical number and classify it.
- For \(g(x)=\ln(x)/x^2\), find and classify every critical number in its domain.
- Why is a zero of a derivative expression not a critical number if it lies outside the original function’s domain?
- For \(h(x)=\ln(4-x^2)\), state the domain. How many separate domain intervals are there?
- When differentiating \(a^{u(x)}\), what role does \(\ln(a)\) play in determining the derivative’s sign?
