Derivative Bounds

How Derivative Bounds Control Function Changes

A derivative describes change near a point. A bound on the derivative can give more: it can control the change in the function across an entire interval. For instance, if the derivative never exceeds a constant $M$, then the function cannot rise by more than $M$ times the distance between two points. Such estimates are useful even when an exact formula for the function’s change is difficult to find.

The Higher Derivatives tutorial established that derivatives can themselves be differentiated. Here we use the first derivative to bound changes in the function, and later apply the same idea to the second derivative. The main argument uses continuity, compactness, and the behavior of a derivative at a local extremum. It does not require the Mean Value Theorem.

A Nonnegative Derivative Gives Monotonicity

We first establish the sign principle that underlies the estimates. A function is nondecreasing on an interval if $g(x)\leq g(y)$ whenever $x<y$ in that interval. The result below turns a pointwise condition on the derivative into this comparison between any two function values.

Theorem (Nonnegative Derivative Implies Nondecreasing): Let $I$ be an open interval, and let $g:I\to\mathbb{R}$ be differentiable. If $g'(t)\geq 0$ for every $t\in I$, then $g$ is nondecreasing on $I$.

Proof. Fix $x,y\in I$ with $x<y$. Suppose, for contradiction, that $g(y)<g(x)$. Write $d=g(x)-g(y)>0$ and $L=y-x>0$, and set $c=d/(2L)>0$. Define $H(t)=g(t)+ct$ on $[x,y]$. Then

H(y)-H(x)=g(y)-g(x)+c(y-x)=-d+\frac{d}{2}=-\frac{d}{2}<0.

For every $t\in(x,y)$, $H'(t)=g'(t)+c\geq c>0$. Also $H'(x)=g'(x)+c>0$, so by the definition of the derivative there is a point $s\in(x,y)$ sufficiently close to $x$ for which $H(s)>H(x)$. The function $H$ is continuous on $[x,y]$, so the Extreme Value Theorem gives a point where it attains its maximum on that interval. Since $H(s)>H(x)$ and $H(y)<H(x)$, neither endpoint can be a point of maximum. Thus a maximum occurs at an interior point of $(x,y)$. At an interior maximum $t_0$, we have $H'(t_0)>0$ because $H'(t)>0$ throughout $(x,y)$. By the definition of the derivative, $H(t_0+h)>H(t_0)$ for all sufficiently small $h>0$, contradicting maximality. Hence $g(y)\geq g(x)$. Since $x<y$ were arbitrary, $g$ is nondecreasing. $\square$

The small added term $ct$ is important in this proof. It makes the derivative strictly positive, while the assumed drop from $g(x)$ to $g(y)$ ensures that $H$ still ends below where it started. The resulting interior maximum contradicts the positive derivative.

Bounds on Secant Slopes

The monotonicity theorem turns upper and lower bounds on $f'$ into bounds on the slope between two points. The derivative need not be constant; it is enough for the same bounds to hold throughout the interval between the points.

Theorem (Secant-Slope Bounds from Derivative Bounds): Let $I$ be an open interval, and let $f:I\to\mathbb{R}$ be differentiable. Suppose $m\leq f'(t)\leq M$ for every $t\in I$, where $m,M\in\mathbb{R}$. Then, whenever $x<y$ are in $I$, $$ m\leq \frac{f(y)-f(x)}{y-x}\leq M. $$

Proof. Define $u(t)=f(t)-mt$ and $v(t)=Mt-f(t)$. Their derivatives satisfy

u'(t)=f'(t)-m\geq0,\qquad v'(t)=M-f'(t)\geq0.

By the Nonnegative Derivative Implies Nondecreasing Theorem, both $u$ and $v$ are nondecreasing. Therefore $u(x)\leq u(y)$ and $v(x)\leq v(y)$. The first inequality gives $f(y)-f(x)\geq m(y-x)$, and the second gives $f(y)-f(x)\leq M(y-x)$. Because $y-x>0$, dividing both inequalities by $y-x$ proves the result. $\square$

Taking $m=-K$ and $M=K$ gives an especially useful consequence. If $|f'(t)|\leq K$, then the absolute value of every secant slope is at most $K$, or equivalently the function is $K$-Lipschitz on $I$.

Worked Example: Secant Slopes of a Cubic

Let $f(x)=x^3-2x$ on the open interval $I=(-1,1)$. Its derivative is $f'(x)=3x^2-2$. Since $0\leq x^2<1$ on $I$, we have $-2\leq 3x^2-2<1$, and in particular $-2\leq f'(x)\leq1$. The Secant-Slope Bounds Theorem therefore gives, for every $x<y$ in $I$,

-2\leq\frac{f(y)-f(x)}{y-x}\leq1.

For example, take $x=0$ and $y=1/2$. The function values are $f(0)=0$ and $f(1/2)=1/8-1=-7/8$, so the secant slope is

\frac{f(1/2)-f(0)}{1/2-0} =\frac{-7/8}{1/2} =-\frac{7}{4}.

Indeed, $-2\leq-7/4\leq1$. The derivative bound controls this slope without requiring us to locate a point where the derivative equals it.

Worked Example: A Decreasing Reciprocal Function

Let $f(x)=1/(x+2)$ on $I=(-1,3)$. Its derivative is $f'(x)=-1/(x+2)^2$. On this interval, $1<x+2<5$, so $1/25<1/(x+2)^2<1$, and hence $-1<f'(x)<-1/25$. Thus for any $x<y$ in $I$, the secant slope lies between $-1$ and $-1/25$.

For $x=0$ and $y=2$, the endpoint values are $f(0)=1/2$ and $f(2)=1/4$. The slope is

\frac{f(2)-f(0)}{2-0} =\frac{1/4-1/2}{2} =-\frac{1}{8}.

The bound is verified directly: $-1\leq-1/8\leq-1/25$. The negative upper bound also confirms that the function decreases: its secant slopes are all negative.

Absolute Derivative Bounds Give Lipschitz Estimates

A two-sided bound on the derivative controls the size of every increment in the function. This is often more useful than a bound on secant slopes because it gives a direct estimate of the difference between function values.

Corollary (Derivative Bound Implies a Lipschitz Bound): Let $I$ be an open interval, and let $f:I\to\mathbb{R}$ be differentiable. If $|f'(t)|\leq K$ for every $t\in I$, where $K\geq0$, then $$ |f(y)-f(x)|\leq K|y-x| $$ for all $x,y\in I$.

Proof. If $x<y$, apply the Secant-Slope Bounds Theorem with $m=-K$ and $M=K$, then multiply by $y-x>0$. This gives $-K(y-x)\leq f(y)-f(x)\leq K(y-x)$, which is equivalent to the asserted absolute-value bound. If $y<x$, interchange the two points; if $x=y$, both sides are zero. $\square$

Worked Example: A Rational Function with a Global Derivative Bound

Consider $f(x)=x/(1+x^2)$ for $x\in\mathbb{R}$. The Quotient Rule gives

f'(x)=\frac{(1+x^2)-2x^2}{(1+x^2)^2} =\frac{1-x^2}{(1+x^2)^2}.

For every real $x$, $|1-x^2|\leq1+x^2$, because $1-x^2\leq1+x^2$ and $x^2-1\leq1+x^2$. Also $1+x^2\geq1$. It follows that

|f'(x)| =\frac{|1-x^2|}{(1+x^2)^2} \leq\frac{1}{1+x^2} \leq1.

The corollary now shows that $|f(y)-f(x)|\leq|y-x|$ for all real $x,y$. For instance, $f(0)=0$ and $f(1)=1/2$, and the estimate reads $1/2\leq1$, as required. The bound applies to every pair of points, not only to points near a chosen input.

A Second Derivative Bounds Changes in the First

The same reasoning applies to higher derivatives. If $f$ is twice differentiable and $f''$ is bounded, then $f'$ has a bounded rate of change. This is one practical way to quantify how rapidly the slope of a function can vary.

Corollary (Second-Derivative Bound): Let $I$ be an open interval, and suppose $f:I\to\mathbb{R}$ is twice differentiable. If $|f''(t)|\leq K$ for every $t\in I$, where $K\geq0$, then $$ |f'(y)-f'(x)|\leq K|y-x| $$ for all $x,y\in I$.

Proof. The function $f'$ is differentiable on $I$, with derivative $f''$. Apply the Derivative Bound Implies a Lipschitz Bound Corollary to the function $f'$. Its derivative is bounded in absolute value by $K$, so the conclusion follows. $\square$

Worked Example: Bounding the Change in a Derivative

Let $f(x)=x^3$ on $I=(-2,2)$. Then $f'(x)=3x^2$ and $f''(x)=6x$. Since $|6x|<12$ on $I$, the second-derivative corollary gives

|f'(y)-f'(x)|\leq12|y-x|\qquad(x,y\in I).

For $x=-1$ and $y=3/2$, the first derivatives are $f'(-1)=3$ and $f'(3/2)=27/4$. The two sides of the estimate are $15/4$ and $12\cdot\frac{5}{2}=30$, respectively, so $15/4\leq30$. The inequality bounds the change in slope using a bound on the second derivative across the whole interval.

Why the Interval Hypothesis Matters

A derivative bound must hold throughout the interval joining the two points. Knowing only the derivative at the endpoints, or only near one of them, does not justify a bound on the entire function increment. The argument depends on applying the monotonicity theorem on the interval between the points.

The open-interval hypotheses above ensure that the derivatives and the local extremum arguments are available at every point used in the proofs. The estimates also apply to any particular closed segment contained in the open interval. In applications where a function is defined on a closed interval, one can state differentiability on its interior and handle endpoints with the corresponding one-sided conditions; those endpoint details should not be silently assumed.

Derivative bounds provide reliable estimates, but they are generally not exact descriptions of a function’s change. For example, the Lipschitz constant $K$ obtained from $|f'|\leq K$ may be larger than necessary. The bound guarantees a uniform estimate, not that any particular pair of points attains equality. The central technique is to use derivative signs to prove monotonicity, then compare the function with simple linear functions whose slopes are the desired bounds.

Check Your Understanding

Use the derivative bounds and proof techniques in this tutorial to answer the following.

State the conclusion of the Nonnegative Derivative Implies Nondecreasing Theorem.
If $2\leq f'(t)\leq5$ throughout an open interval, what bounds follow for the secant slope between any two ordered points?
Why does a bound $|f'(t)|\leq K$ imply an estimate for every pair of points, rather than only for nearby points?
For $f(x)=x^2$ on $(-1,1)$, find a bound for $|f'(x)|$, and use it to give a Lipschitz estimate on that interval.
If $|f''(t)|\leq4$, what estimate follows for $|f'(y)-f'(x)|$?

What You'll Learn