Turn a Claim Into a Function
The Uniform Convergence Strategy used estimates that controlled errors throughout a domain. Differentiation offers another route to global conclusions: when a function has a derivative of a known sign, the Mean Value Theorem can turn that local information into a comparison between values at different points. This makes derivatives useful not only for calculating rates of change, but also for proving inequalities and uniqueness statements.
The central move is to encode the claim in an auxiliary function. To prove that \(F(x)\geq G(x)\), for example, consider \(H(x)=F(x)-G(x)\). The desired inequality becomes \(H(x)\geq0\). If a derivative calculation shows that \(H\) decreases up to a point where it is zero and increases afterward, then that point is a global minimum and the inequality follows. The comparisons must be made on intervals that actually include the points being compared.
In the strategy developed here, the derivative is usually not the conclusion. It is evidence for a comparison. The Mean Value Theorem is the bridge: if \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then for some \(c\in(a,b)\),
Thus, when \(a<b\), the sign of \(f'(c)\) determines the sign of the difference \(f(b)-f(a)\). The hypotheses matter: continuity on the closed interval and differentiability in its interior are what permit this conclusion.
From a Derivative Sign to Monotonicity
A useful first result makes the derivative-to-comparison step precise. It is often called the derivative sign criterion for monotonicity. The non-strict and strict versions have different hypotheses: a nonnegative derivative guarantees nondecrease, while a positive derivative guarantees strict increase.
Proof. Take any \(a,b\in I\) with \(a<b\). Since \(I\) is an interval, \([a,b]\subseteq I\). The hypotheses ensure that \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\). By the Mean Value Theorem, there is a \(c\in(a,b)\) such that
Because \(b-a>0\), if \(f'\geq0\) throughout the interior, then \(f(b)-f(a)\geq0\), so \(f(a)\leq f(b)\). This holds for every \(a<b\), proving that \(f\) is nondecreasing. If \(f'>0\) throughout the interior, the same equation gives \(f(b)-f(a)>0\), proving strict increase. If \(f'\leq0\), then \(f(b)-f(a)\leq0\); if \(f'<0\), then \(f(b)-f(a)<0\). These yield the two decreasing versions. \(\square\)
The theorem compares points in the interval, including its endpoints when they are present. It does not say that knowing a derivative sign on an open interval automatically compares a point inside that interval with a boundary point outside it. For such a comparison, apply the theorem on a closed interval containing both points, or apply the Mean Value Theorem directly there.
Encode an Inequality as a Minimum
A common differentiation proof starts by moving all terms of an inequality to one side. Then one studies the resulting function, locates a point where its derivative changes sign, and compares its values with the value at that point. The intervals used in those comparisons must include the point where equality is claimed.
Worked Example: Prove \(\ln x\leq x-1\)
We prove the inequality for every \(x>0\). Define
The desired claim is \(g(x)\geq0\). Since \(g(1)=0\), it is enough to show that \(g(x)\geq g(1)\) for every positive \(x\). Differentiation gives
For \(0<x<1\), apply the Mean Value Theorem on \([x,1]\). The function \(g\) is continuous there and differentiable on \((x,1)\). For the point \(c\in(x,1)\) supplied by the theorem, \(g'(c)<0\), and \(1-x>0\). Therefore
so \(g(x)>g(1)=0\). For \(x>1\), apply the theorem instead on \([1,x]\). Its intermediate point \(c\in(1,x)\) satisfies \(g'(c)>0\), and \(x-1>0\). Hence
so again \(g(x)>g(1)=0\). At \(x=1\), equality holds. Thus \(g(x)\geq0\) for every \(x>0\), which is exactly \(\ln x\leq x-1\).
The endpoint choices are essential to the argument. Strict decrease on \((0,1)\) alone compares two points both in \((0,1)\); it does not by itself compare \(g(x)\) with \(g(1)\). Applying the Mean Value Theorem on \([x,1]\) supplies that comparison directly. The same care is needed on the other side of \(1\).
Worked Example: Bound \(\ln(1+x)\) on the Nonnegative Axis
For every \(x\geq0\), we will prove the two-sided estimate
For the upper bound, define \(u(x)=x-\ln(1+x)\) on \([0,\infty)\). We have \(u(0)=0\), and for \(x>0\),
The derivative sign criterion shows that \(u\) is increasing on \([0,\infty)\). Consequently \(u(x)\geq u(0)=0\), or \(\ln(1+x)\leq x\). At \(x=0\), this is equality.
For the lower bound, define \(v(x)=\ln(1+x)-x/(1+x)\). Again \(v(0)=0\). For \(x>0\), the quotient rule gives
The denominator is positive for \(x\geq0\), so the derivative sign criterion shows that \(v\) is increasing and \(v(x)\geq v(0)=0\). This is the lower bound. Both inequalities hold at \(x=0\) as well, and together they prove the estimate on the whole nonnegative axis.
Use Monotonicity to Prove Uniqueness
The same strategy applies when the goal is not an inequality but uniqueness. If a function is strictly increasing, it cannot take the same value at two different points. Therefore, to prove an equation has at most one solution, define a function whose zeros are exactly the solutions and show that its derivative is strictly positive or strictly negative on the domain.
Worked Example: Uniqueness of a Logarithmic Equation
Consider the equation
Define \(q(x)=2x+\ln x-2\) on \((0,\infty)\). A solution of the equation is exactly a zero of \(q\). For \(x>0\),
By the derivative sign criterion, \(q\) is strictly increasing on \((0,\infty)\). A strictly increasing function has at most one zero: if \(a<b\) were both zeros, strict increase would imply \(q(a)<q(b)\), contradicting \(q(a)=q(b)=0\). Finally, \(q(1)=2+\ln 1-2=0\), so \(x=1\) is a solution. It follows that \(x=1\) is the unique solution.
This proof separates two tasks that are easy to conflate. Strict monotonicity proves at most one solution. Substitution of \(x=1\) proves at least one solution. Together they establish existence and uniqueness.
A Derivative Proof of a Tangent-Line Bound
Auxiliary functions also compare a curved function with a simpler expression. The next example uses a derivative estimate to show that a square-root graph lies below a particular line. The equality point is where the difference between the line and the function is zero.
Worked Example: A Square-Root Inequality
For \(x\geq0\), prove
Set \(r(x)=1+x/2-\sqrt{1+x}\). The claim is \(r(x)\geq0\), and \(r(0)=0\). For \(x>0\),
The denominator is positive, and \(\sqrt{1+x}>1\) when \(x>0\), so the derivative is strictly positive. By the derivative sign criterion, \(r\) is increasing on \([0,\infty)\). Therefore \(r(x)\geq r(0)=0\), which proves the inequality; equality occurs at \(x=0\).
As a direct algebraic check, both sides of the claimed inequality are nonnegative on this domain. Squaring them is therefore legitimate, and gives the equivalent comparison \(1+x\leq(1+x/2)^2=1+x+x^2/4\), which holds because \(x^2/4\geq0\). The differentiation proof illustrates how to establish the comparison by locating the minimum of the difference function.
Choose the Interval Before Drawing the Conclusion
Differentiation is powerful because the Mean Value Theorem converts information about \(f'\) into information about differences \(f(b)-f(a)\). But a derivative calculation by itself is not yet a proof of a global claim. The interval, endpoint values, and hypotheses must all fit together.
- For an inequality, define a difference function so that the desired statement becomes a sign condition.
- Find where the derivative is positive, negative, or zero, and use the derivative sign criterion only on an interval where its hypotheses hold.
- Compare the relevant points on a closed interval that contains them. Do not use a monotonicity statement on an open interval to compare with a boundary point it excludes.
- For uniqueness, prove strict monotonicity on the entire solution domain, then check existence separately if the claim requires it.
A derivative changing sign from negative to positive suggests a minimum; one changing from positive to negative suggests a maximum. To turn that suggestion into a proof, split the domain at the critical point and compare each point with it using the Mean Value Theorem or the derivative sign criterion. A zero derivative at one point alone does not establish an extremum, and derivative information on only part of the domain does not establish a global comparison.
For an inequality, subtract one side from the other. For an equation, define a function whose zeros are precisely its solutions.
Check the domain and verify that every denominator or other sign-sensitive expression has the required sign.
Ensure the interval contains both points being compared and that continuity and differentiability hold as required.
Use a known function value to prove an inequality or existence; use strict monotonicity to rule out two distinct solutions.
The differentiation strategy is most reliable when each step has a clear role: the auxiliary function translates the claim, the derivative reveals monotonicity, and the Mean Value Theorem justifies the comparison. Careful interval choices ensure that the conclusion reaches the points the proof actually needs.
Check Your Understanding
Use the derivative sign criterion and the worked proof strategies to answer the following questions.
- Why does \(f'(x)\geq0\) on an interval imply that \(f\) is nondecreasing there?
- In the proof of \(\ln x\leq x-1\), why is the interval \([x,1]\) used when \(0<x<1\)?
- What function would you define to prove an inequality \(F(x)\geq G(x)\) by differentiation?
- Why does strict monotonicity prove at most one solution, but not by itself prove that a solution exists?
- Which continuity and differentiability hypotheses are needed to apply the Mean Value Theorem on \([a,b]\)?