Constant Rule: Why The Derivative Of A Constant Is 0

What Is The Constant Rule?

The constant rule states that the derivative of a constant function is zero: if $f(x) = c$ for a fixed real number $c$, then $\frac{d}{dx}f(x) = 0$ for every $x$. A constant function outputs the same value no matter what you feed it, so its output never rises or falls, and a quantity that never changes has a rate of change of zero.

In symbols:

$$\frac{d}{dx},(c) = 0$$

Here $c$ is any fixed real number (it can be $7$, $-325$, $\pi^5$ or $\ln 3$), $x$ is the input variable, and $\frac{d}{dx}$ means "the rate of change with respect to $x$". The rule holds at every real $x$ with no extra conditions, because a constant function is defined and perfectly smooth everywhere.

The constant rule is the first entry in every derivative table, and it quietly does work inside almost every later calculation. Differentiating $4x^3 - 9x + 12$ needs it for the $12$. Finding a velocity from a height formula needs it for the starting height. Each time a fixed number sits beside the variable, the constant rule is what deletes it.

How Do You Prove The Constant Rule From The Limit Definition?

The proof takes three lines once you write out the definition of the derivative. A reader on Math Stack Exchange asked for a "proof that the derivative of a constant is zero" after finding only a version that rewrote $c$ as $c \cdot x^0$; the cleaner route starts from the definition itself, the same way OpenStax Calculus Volume 1 proves it. A Quora thread asks the same thing in plain words: can you prove that the derivative of any constant or number is zero by the first principle? The lines below are that first-principle proof.

The limit definition of the derivative says

$$\frac{d}{dx}f(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.$$

Take $f(x) = c$. Then $f(x+h) = c$ as well, since the function ignores its input:

$$\frac{d}{dx},(c) = \lim_{h \to 0} \frac{c - c}{h}$$

$$= \lim_{h \to 0} \frac{0}{h}$$

$$= \lim_{h \to 0} 0$$

$$= 0$$

The step from $\frac{0}{h}$ to $0$ is legitimate because $h$ approaches $0$ without ever equalling it, so $\frac{0}{h}$ is exactly $0$ for every $h$ in play. The last step uses the simplest of the limit laws: the limit of a constant is that constant, here the constant $0$.

What Does The Constant Rule Mean Geometrically?

The graph of $y = c$ is a horizontal line, and a horizontal line has slope zero at every point. Since the derivative at a point is the slope of the tangent line there, the derivative of a constant is zero everywhere.

The tangent line to $y = c$ at any point is the line itself. Ask for the tangent at $x = 2$ on the graph of $y = 5$ and the answer is $y = 5$, the same horizontal line. That is an unusual situation in calculus: for most curves the tangent touches at one point and then pulls away, but a flat line already is its own best straight-line approximation. Every tangent line is built from a slope and a point, and here the slope is $0$.

There is a second geometric reading that matters more later. Add a constant to any function and the graph slides straight up or down without tilting, the vertical shift described under transformations of functions. Every point moves the same distance, so every tangent line keeps its slope. Algebraically, $\frac{d}{dx}\left[f(x) + c\right] = \frac{d}{dx}f(x) + 0$.

Can You Use The Power Rule To Get The Constant Rule?

Partly, and only with a gap at $x = 0$. A Math Stack Exchange question asks exactly this: "Can I apply power rule to the derivative of constant function?" by writing $k = k \cdot x^0$.

The argument goes like this. The power rule says $\frac{d}{dx}x^n = n,x^{n-1}$. With $n = 0$:

$$\frac{d}{dx}\left(k \cdot x^0\right) = k \cdot 0 \cdot x^{-1} = 0$$

The trouble is that $x^{-1} = \frac{1}{x}$ does not exist at $x = 0$, and $x^0$ itself is undefined there under many conventions. So the power-rule route proves the result only for $x \neq 0$, while the limit proof covers every real $x$ in one stroke. The cleaner way to think about it: the constant rule is the more basic fact, and the power rule's $n = 0$ case agrees with it wherever both make sense.

The same habit of separating "a number" from "the variable" explains why $\frac{d}{dx}(x) = 1$ while $\frac{d}{dx}(5) = 0$. The variable $x$ moves one unit for every unit of input; the number $5$ does not move at all.

Is The Constant Rule The Same As The Constant Multiple Rule?

No. The constant rule handles a constant standing alone, and the result is always $0$. The constant multiple rule handles a constant multiplying a function, and the constant survives: $\frac{d}{dx}\left[c,f(x)\right] = c,\frac{d}{dx}f(x)$.

The two rules look alike on the page, which is exactly why they get swapped. The deciding question is whether the constant is added or multiplied:

A quick test settles any case: if deleting the constant changes the shape of the graph's slopes, it was multiplying; if deleting it only slides the graph, it was added.

Which Quantities Count As Constants?

A constant is any quantity that does not depend on the variable you are differentiating with respect to. That covers plain numbers, but it also covers expressions that look busy:

The last row is the one to watch. Whether something is a constant depends on the variable. In $y = a^2 + a,x$, differentiating with respect to $x$ treats $a$ as fixed, so $\frac{dy}{dx} = a$. Differentiating the same expression with respect to $a$ treats $x$ as fixed, so $\frac{dy}{da} = 2a + x$.

The opposite trap shows up in implicit differentiation. There, $y$ is a function of $x$ even though no formula for it is written, so $\frac{d}{dx}\left(y^2\right) = 2y,\frac{dy}{dx}$, not $0$. A letter is a constant only if it truly stays fixed while $x$ moves.

Examples Of The Constant Rule

Example 1

Find $\frac{d}{dx},(7)$ and $\frac{d}{dt}\left(-\frac{3}{4}\right)$.

Both functions are constants, so the constant rule applies directly.

$$\frac{d}{dx},(7) = 0$$

$$\frac{d}{dt}\left(-\frac{3}{4}\right) = 0$$

The sign and the fraction make no difference, and neither does the name of the variable.

Final answer: both derivatives equal $0$.

Example 2

Find $\frac{dy}{dx}$ for $y = \pi^5$.

This one appeared on a math T-shirt and then on Math Stack Exchange under the title "Derivative of $\pi^5$", where the asker had found three different answers.

The tempting path treats $\pi$ like a variable and applies the power rule:

$$\frac{d}{dx}\left(\pi^5\right) = 5\pi^4 \approx 487.0455$$

Check that answer against the graph. The number $\pi^5$ is about $306.0197$, so $y = \pi^5$ is the horizontal line $y \approx 306.0197$. A horizontal line has slope $0$, yet the power-rule answer claims a slope of about $487$. Something has gone wrong: the power rule differentiates powers of the variable, and $\pi$ is not the variable.

The correct route asks what $y$ depends on. Nothing in $\pi^5$ involves $x$, so it is a constant:

$$\frac{dy}{dx} = \frac{d}{dx}\left(\pi^5\right)$$

$$= 0$$

Final answer: $\frac{dy}{dx} = 0$.

Example 3

Differentiate $f(x) = 4x^3 - 9x + 12$, then differentiate $g(x) = 4x^3 - 9x + 62$.

Differentiate term by term. The constant rule removes the constant term.

$$\frac{d}{dx}f(x) = \frac{d}{dx}\left(4x^3\right) - \frac{d}{dx},(9x) + \frac{d}{dx},(12)$$

$$= 12x^2 - 9 + 0$$

$$= 12x^2 - 9$$

For $g$, only the constant changed, from $12$ to $62$:

$$\frac{d}{dx}g(x) = 12x^2 - 9 + 0 = 12x^2 - 9$$

The two functions differ by $50$ everywhere, so their graphs are vertical shifts of each other and share every slope.

Final answer: both derivatives equal $12x^2 - 9$.

Example 4

A ball is thrown upward from a 20 m balcony. Its height after $t$ seconds is $s(t) = 20 + 15t - 4.9t^2$ metres. Find its velocity at $t = 1$ second, and compare with a ball thrown the same way from the ground.

Velocity is the derivative of height with respect to time.

$$\frac{ds}{dt} = \frac{d}{dt},(20) + \frac{d}{dt},(15t) - \frac{d}{dt}\left(4.9t^2\right)$$

$$= 0 + 15 - 9.8t$$

At $t = 1$:

$$\frac{ds}{dt} = 15 - 9.8 = 5.2$$

From the ground, $s(t) = 15t - 4.9t^2$, and the derivative is again $15 - 9.8t$. The starting height of $20$ m changes where the ball is, not how fast it moves.

Final answer: $5.2,\text{m/s}$ upward at $t = 1$, the same from the balcony and from the ground.

Example 5

Evaluate $\int 6,dx$ and $\int_{2}^{5} 6,dx$.

Run the constant rule backwards. Which function has derivative $6$? The function $6x$ does, since $\frac{d}{dx},(6x) = 6$, and adding any constant keeps that true:

$$\int 6,dx = 6x + C$$

For the definite integral, use the antiderivative at the two bounds:

$$\int_{2}^{5} 6,dx = 6(5) - 6(2)$$

$$= 30 - 12$$

$$= 18$$

Geometrically, the region under $y = 6$ from $x = 2$ to $x = 5$ is a rectangle of height $6$ and width $3$, with area $18$.

Final answer: $\int 6,dx = 6x + C$ and $\int_{2}^{5} 6,dx = 18$.

Example 6

A function $f$ satisfies $\frac{df}{dx} = 0$ for every real $x$, and $f(2) = 5$. Find $f(x)$ and the tangent line at $x = 2$.

A function whose derivative is zero on an entire interval must be constant there (this is the converse of the constant rule, covered in the next section). The whole real line is an interval, so $f(x) = c$ for some $c$.

$$f(2) = c = 5$$

$$f(x) = 5$$

The tangent line at $x = 2$ has slope $0$ and passes through $(2, 5)$:

$$y - 5 = 0 \cdot (x - 2)$$

$$y = 5$$

Final answer: $f(x) = 5$, and the tangent line at $x = 2$ is $y = 5$, the graph itself.

If The Derivative Is Zero, Must The Function Be Constant?

Yes, on an interval. If $\frac{df}{dx} = 0$ at every point of an open interval, then $f$ is constant on that interval. This converse is a corollary of the mean value theorem: for any two points $a < b$ in the interval, there is some $\xi$ between them with

$$f(b) - f(a) = \frac{df}{dx}\bigg|_{x = \xi} \cdot (b - a) = 0 \cdot (b - a) = 0,$$

so $f(b) = f(a)$ for every pair of points.

The interval condition is not decoration; Wikipedia's entry on the constant function states the converse for functions defined on all real numbers, an interval. The function

$$f(x) = \frac{x}{|x|}, \qquad x \neq 0,$$

equals $-1$ for every negative $x$ and $1$ for every positive $x$. Its derivative is $0$ at every point of its domain, yet it is not constant, because the domain has a gap at $0$ and the mean value theorem cannot reach across a gap. This took me a moment the first time it came up; the derivative test is local, and "constant" is a statement about the whole domain.

The converse is also the reason antiderivatives come with $+,C$. Two functions with the same derivative differ by a function whose derivative is zero, and on an interval that difference is a constant.

What Is The Integral Of A Constant?

The integral of a constant $k$ is $kx + C$, not $k + C$. Integration asks for a function whose derivative is $k$, and $\frac{d}{dx},(kx) = k$:

$$\int k,dx = kx + C$$

Here $C$ is the constant of integration, which appears precisely because the constant rule erases constants when you differentiate: $kx + 3$, $kx - 10$ and $kx + \pi$ all have derivative $k$, so the integral must allow every one of them. That is the rule's two sides in one line: differentiation sends a constant to $0$, and every indefinite integral has to put back an unknown constant.

On a closed interval the picture is a rectangle. The region under the horizontal line $y = k$ (for $k > 0$) from $x = a$ to $x = b$ has height $k$ and width $b - a$:

$$\int_{a}^{b} k,dx = k,(b - a)$$

Why Does The Constant Rule Matter?

The constant rule exists because rates do not depend on where you start measuring. A rate of change compares two readings, and any fixed offset added to both readings cancels in the subtraction. Three everyday places make this concrete:

• Temperature scales. Kelvin and Celsius differ by a fixed $273.15$, and NIST notes that one Celsius degree is an interval of 1 K. So $\frac{d}{dt}\left(T + 273.15\right) = \frac{dT}{dt}$, and a weather model can report warming rates in either unit without conversion.

• Height and position. Measuring height from sea level instead of from the floor adds a constant, and velocity, the derivative of height, ignores it. Example 4 showed the balcony and the ground giving the same $5.2,\text{m/s}$.

• Cost. A factory's fixed costs (rent, machines) are a constant in the total cost function, so they vanish from the marginal cost. With $C(q) = 500 + 12q$, the cost of one more unit is $\frac{dC}{dq} = 12$, whatever the rent; see marginal cost and marginal revenue.

This offset-blindness is the whole point. It is what lets engineers, scientists and economists choose any convenient zero for their measurements, knowing the rates they compute will not change.

Where Do Students Trip Up On The Constant Rule?

Most errors with the constant rule are not about the rule itself. They come from applying it to something that is not a lone constant, or from carrying it into integration in the wrong direction.

Mistake 1: Zeroing Out A Constant That Multiplies

Where it slips in:

Differentiating terms like $5x$ or $-3x^2$ right after learning that "constants go to zero".

Don't do this:

Writing $\frac{d}{dx},(5x) = 0$ because the $5$ is a constant.

The correct way:

A constant multiplying the variable stays: $\frac{d}{dx},(5x) = 5$. The constant rule applies only to a constant standing alone.

Students who have just learned the rule tend to zero out every number they see, including the $5$ in $5x$ and the $4$ in $4x^3$; the fix is to ask "added or multiplied?" before touching any number. The rusher makes this error most often, because the shortcut "numbers disappear" feels finished before the term has been read.

Mistake 2: Integrating A Constant To Itself

Where it slips in:

Moving from derivatives to integrals, as in the Math Stack Exchange question "Indefinite integral of a constant", which asks whether the answer is $k + C$ or $kx + C$.

Don't do this:

Writing $\int 9,dx = 9 + C$.

The correct way:

Check by differentiating: $\frac{d}{dx},(9 + C) = 0$, not $9$. The right answer is $\int 9,dx = 9x + C$, since $\frac{d}{dx},(9x + C) = 9$.

The memorizer is the usual victim here: "constants vanish" was stored as a fact about constants, not as a fact about differentiating them, so it gets applied in both directions. The habit that fixes it is the derivative check, which takes five seconds and catches every integral of this kind.

Mistake 3: Treating A Moving Letter As A Constant

Where it slips in:

Implicit differentiation, or any formula where $y$ depends on $x$ without a written formula.

Don't do this:

Writing $\frac{d}{dx}\left(y^2\right) = 0$ because $y$ "is not $x$".

The correct way:

Ask whether the letter changes when $x$ changes. If it does, use the chain rule: $\frac{d}{dx}\left(y^2\right) = 2y,\frac{dy}{dx}$. Only letters that stay fixed, like a parameter $a$ or the number $\pi$, differentiate to $0$.

Mistake 4: Assuming Zero Derivative Means Constant Everywhere

Where it slips in:

Functions with a split domain, such as $\frac{x}{|x|}$ or functions defined piecewise on separate intervals.

Don't do this:

Concluding $f$ is a single constant just because $\frac{df}{dx} = 0$ wherever the derivative exists.

The correct way:

State the interval. Zero derivative gives a constant on each connected interval of the domain, and the constants can differ from piece to piece.

Practice Problems On The Constant Rule

Try each problem before reading its answer.

1. Find $\frac{d}{dx},(-19)$.

2. Find $\frac{d}{dx}\left(\ln 7 + x^2\right)$.

3. Find $\frac{d}{dx}\left(3x^4 - 2x + \sqrt{5}\right)$.

4. Evaluate $\int 9,dx$.

5. Evaluate $\int_{-1}^{3} 4,dx$.

6. A function $f$ has $\frac{df}{dx} = 0$ on the interval $(0, 10)$ and $f(1) = -2$. Find $f(7)$.

Answer to Problem 1: $0$.

Answer to Problem 2: $2x$, because $\ln 7$ is a constant.

Answer to Problem 3: $12x^3 - 2$, because $\sqrt{5}$ is a constant.

Answer to Problem 4: $9x + C$.

Answer to Problem 5: $16$, the area of a rectangle of height $4$ and width $4$.

Answer to Problem 6: $-2$, because $f$ is constant on the interval $(0, 10)$, which contains both $1$ and $7$.

Key Takeaways

• The constant rule says the derivative of any constant is $0$, because a constant never changes.

• The limit definition proves it in three lines, since $\frac{c - c}{h} = 0$ for every $h \neq 0$.

• A constant added to a function disappears when you differentiate, while a constant multiplying a function stays (the constant multiple rule).

• The integral of a constant $k$ is $kx + C$, and on an interval it equals a rectangle's area $k,(b - a)$.

• A zero derivative means constant only on an interval, which is also why every antiderivative carries $+,C$.

Practice These Before Moving On

Work through Practice Problems 1 to 6, then redo Examples 2 and 3 without looking; if a constant multiplying $x$ keeps turning into $0$, return to the comparison table in "Is The Constant Rule The Same As The Constant Multiple Rule?", and if an integral of a constant comes out as $k + C$, return to "What Is The Integral Of A Constant?".

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Read More

• Table of derivatives for the common functions

• Basic integration formulas, including constants and powers

• The derivative of a function and what it measures

• How differentiation and integration undo each other

• Antiderivatives and families of solutions