Why Constant Factors Matter
A function may describe a changing quantity in one scale, while a constant multiple of that function describes the same variation in another scale. For example, if \(f(x)\) records a quantity, then \(cf(x)\) rescales its values by the fixed factor \(c\). Its rate of change is rescaled by that same factor.
This observation is a basic differentiation rule. It applies whether the constant is positive, negative, fractional, or zero. The difference quotient explains why: the constant multiplies every change in the function, so it multiplies the limiting rate as well.
Definition and Statement of the Rule
Let \(f\) be defined on an open interval containing \(a\). The derivative of \(f\) at \(a\), when it exists, is the limit
The number \(h\) is restricted so that \(a+h\) remains in the interval. The function is differentiable at \(a\) if this limit exists as a finite real number.
The first statement is pointwise: it concerns one input \(a\) at which differentiability is assumed. The second says the rule holds at every input where \(f\) is differentiable; it does not assert that \(f\) is differentiable everywhere in its domain.
Why the Rule Works: A Difference-Quotient Proof
Set \(g(x)=cf(x)\). At an input \(a\), the difference quotient for \(g\) is
Since \(c\) is the same fixed number in both terms, factor it out of the numerator and then out of the quotient:
By assumption, the quotient for \(f\) tends to the finite limit \(f'(a)\) as \(h\to 0\). A constant multiple of a function with a finite limit has the limit equal to that constant times the original limit. Therefore,
This proves both that \(g\) is differentiable at \(a\) and that its derivative there is \(cf'(a)\). The argument does not require \(c\) to be positive. In particular, a negative factor reverses the sign of the derivative, a fractional factor scales its size, and \(c=0\) gives the derivative of the zero function.
Worked Examples
Worked Example: A Negative Fractional Factor
Differentiate \(g(x)=-\frac{3}{2}x^2\), and check the result at \(x=2\) using the difference quotient.
Take \(f(x)=x^2\) and \(c=-\frac{3}{2}\). The derivative of \(f\) is \(f'(x)=2x\). Applying the constant multiple rule gives
At \(x=2\), this predicts \(g'(2)=-3(2)=-6\). Check directly from the definition. For \(h\ne 0\),
Taking \(h\to 0\) gives \(-6\), which agrees with the rule. Thus the value from the formula is checked directly at the specified input.
Worked Example: A Fractional Factor with a Root
Let \(g(x)=\frac{1}{3}\sqrt{x}\), defined for \(x\ge 0\). Find \(g'(4)\) and verify it from the difference quotient.
The function \(f(x)=\sqrt{x}\) is differentiable at \(4\), with \(f'(x)=\frac{1}{2\sqrt{x}}\) for \(x>0\). The rule gives
To check, for nonzero \(h\) sufficiently close to zero,
As \(h\to 0\), the denominator tends to \(3(2+2)=12\), so the quotient tends to \(\frac{1}{12}\). This confirms the value from the rule. The input \(4\) is an interior point of the domain, so the two-sided derivative is appropriate.
Worked Example: A Negative Factor with a Reciprocal
Let \(g(x)=-\frac{2}{5x}\), for \(x\ne 0\). Find \(g'(1)\) and verify it using the difference quotient.
Write \(g(x)=-\frac{2}{5}f(x)\), where \(f(x)=\frac{1}{x}\). Since \(f'(x)=-\frac{1}{x^2}\) for \(x\ne 0\),
For the direct check, take \(h\ne 0\) close enough to zero that \(1+h\ne 0\). Then
The limit as \(h\to 0\) is \(\frac{2}{5}\), matching the rule. In this example, the derivative of \(f\) is negative at \(1\); multiplying it by the negative constant makes the derivative of \(g\) positive.
Using the Rule Reliably
To apply the constant multiple rule at a particular input, first identify the fixed factor \(c\) and the function \(f\). Confirm that \(f\) is differentiable at the input. Then calculate \(f'\) there and multiply by \(c\). The constant is not differentiated separately: it is a fixed scale factor, not a changing function of the input.
Write the expression as \(cf(x)\), with \(c\) independent of \(x\).
Use the rule only at inputs where \(f\) is differentiable.
Compute \(f'(x)\), then multiply by \(c\), preserving its sign and magnitude.
At a specified input, compare with the difference quotient or substitute that input into the derivative formula.
The value \(c=0\) also fits the rule. If \(g(x)=0f(x)\), then \(g(x)=0\) wherever \(f\) is defined, so its difference quotient is \(0\) and its derivative is \(0\). The formula gives \(0f'(a)=0\) at every point where \(f\) is differentiable. The rule makes no claim at a point where the original function is not differentiable.
A Common Sign Mistake
A frequent error is to lose a negative constant when multiplying the derivative. For example, consider \(q(x)=-\frac{1}{2}\sqrt{x}\). A wrong calculation is
The last equality is false: the product of a negative number and a positive number is negative. The corrected calculation is
At \(x=4\), the corrected value is \(-\frac{1}{8}\), whereas the incorrect expression gives \(\frac{1}{8}\). A difference-quotient check confirms the negative sign:
Keep the constant attached to the derivative until the multiplication is complete. This makes its sign harder to lose and helps distinguish the constant multiple rule from the incorrect idea that a constant factor makes the derivative vanish.
Why the Rule Is Useful
The rule preserves the scale of rates. If \(f\) measures a quantity and \(cf\) expresses that quantity in a scale multiplied by \(c\), then the rate of change is multiplied by \(c\) too. A negative \(c\) reverses the sign of the rate, while a fractional \(c\) reduces its magnitude according to the same factor. The rule therefore connects algebraic scaling of function values with scaling of their instantaneous rates.
Its conditions matter just as much as its formula. The constant must be independent of the input, and the derivative conclusion is available only where the original function is differentiable. Keeping these conditions explicit prevents the rule from being used to claim a derivative exists where it does not.
Check Your Understanding
Answer these questions using the definition and rule developed above.
- If \(f\) is differentiable at \(a\), what is the derivative of \(-7f\) at \(a\)?
- In the difference-quotient proof, where does the constant \(c\) appear after factoring?
- Suppose \(f'(3)=-4\). What is \(\left(\frac{1}{2}f\right)'(3)\)?
- Why does the rule not assert that \(cf\) has a derivative at every point in the domain of \(f\)?
- Find the sign of the derivative of \(-\frac{1}{4}f\) at a point where \(f'\) is positive. Explain how the sign follows from the rule.
