Variable rule

The variable changes exactly as fast as itself.

\[\frac{d}{dx}x = 1\]

What it says

If \(f(x) = x\), then increasing \(x\) by \(h\) increases the value by exactly \(h\). The difference quotient is \(\frac{(x + h) - x}{h} = \frac{h}{h} = 1\) for every \(h \neq 0\), so the derivative is \(1\).

This is the base case of the whole process. Every other rule reduces a complicated expression to the derivatives of simpler pieces; the recursion has to stop somewhere, and it stops here and at the constant rule.

It is also the \(n = 1\) case of the power rule: \(\frac{d}{dx}x^{1} = 1 \cdot x^{0} = 1\).

When it applies

  • The expression is the differentiation variable standing on its own.
  • At the bottom of a chain, product or quotient computation, once everything else has been peeled away.
  • Any other symbol is not the variable: with respect to \(x\), the derivative of \(y\) is \(0\).

Five worked examples

Every line is the step the calculator would show, in the order it applies them. Each graph is live: hover it to read both curves and see the tangent whose slope is the derivative, drag to pan, scroll to zoom.

Example 1
\[\frac{d}{dx}\left[x\right]\]
  1. The expression is the variable itself.

    \[\frac{d}{dx}\left[x\right] = 1\]
Answer
\[1\]

f is the 45° line and f′ is the constant 1. Read the pair vertically — the height of f′ at any point is the steepness of f directly above it, and that steepness never changes. Hover anywhere and f′ reads exactly 1.

The line \(y = x\) rises one unit for every unit across — slope \(1\).

Open x in the calculator
Example 2
\[\frac{d}{dt}\left[t\right]\]
  1. Here the differentiation variable is \(t\), and the expression is \(t\).

    \[\frac{d}{dt}\left[t\right] = 1\]
Answer
\[1\]

The identical picture with t on the axis instead of x. The letter is irrelevant; the expression matches the variable being differentiated.

The letter does not matter, only whether it matches the variable you are differentiating with respect to.

Open t in the calculator
Example 3
\[\frac{d}{dx}\left[x + 5\right]\]
  1. Split the sum.

    \[\frac{d}{dx}\left[x + 5\right] = \frac{d}{dx}\left[x\right] + \frac{d}{dx}\left[5\right]\]
  2. The variable rule gives \(1\); the constant rule gives \(0\).

    \[= 1 + 0 = 1\]
Answer
\[1\]

The 45° line lifted so that it crosses the axis at \(x = -5\), with f′ still pinned at 1 — the two lines are parallel everywhere.

Adding a constant shifts the graph up or down without tilting it, so the slope is unchanged.

Open x + 5 in the calculator
Example 4
\[\frac{d}{dx}\left[5 x\right]\]
  1. Pull the constant factor out.

    \[\frac{d}{dx}\left[5x\right] = 5\;\frac{d}{dx}\left[x\right]\]
  2. Then the variable rule.

    \[= 5 \cdot 1 = 5\]
Answer
\[5\]

Five times the steepness, and f′ answers with the constant 5. Its height has changed while its shape has not.

The \(1\) is where the answer comes from, even though it is invisible in the final \(5\).

Open 5x in the calculator
Example 5
\[\frac{d}{dx}\left[x \sin\left(x\right)\right]\]
  1. Product rule, with \(f = x\) and \(g = \sin x\).

    \[\frac{d}{dx}\left[x\sin x\right] = \frac{d}{dx}\left[x\right]\sin x + x\,\frac{d}{dx}\left[\sin x\right]\]
  2. The variable rule supplies the first derivative, the trigonometric rule the second.

    \[= 1 \cdot \sin x + x\cos x\]
Answer
\[\sin\left(x\right) + x \cos\left(x\right)\]

Here the variable rule is only one factor inside a product: f oscillates with a growing swing, and f′ crosses zero at each crest and trough.

In real problems the variable rule almost always appears as a factor of \(1\) inside a bigger rule.

Open x sin(x) in the calculator

Watch out for

  • The derivative of \(x\) is \(1\), not \(x\) and not \(0\).
  • With respect to a different variable, \(x\) is a constant: \(\frac{d}{dt}\left[x\right] = 0\).
  • The \(1\) usually disappears into a product, so it is easy to forget it was ever applied.