What a Derivative’s Sign Tells You
A derivative describes how an output changes as its input changes. Its sign tells whether the output is locally higher or lower when the input moves forward from a specified value. The interpretation applies in many settings: position changing with time, temperature changing with time, or cost changing with time.
The sign must be interpreted at a particular input. A positive derivative at one input does not, by itself, tell us what happens throughout a much larger interval. It tells us about output changes close to that input, and the difference quotient makes the precise meaning clear.
From the Difference Quotient to a Local Comparison
Suppose \(f\) is differentiable at \(a\). By definition, its derivative there is the limit of the difference quotient:
For \(h>0\), the denominator is positive. Therefore the sign of the quotient is the sign of the output change \(f(a+h)-f(a)\). If the derivative is positive, the quotient is positive for sufficiently small \(h>0\), so the output at \(a+h\) is greater than the output at \(a\). If the derivative is negative, the quotient is negative, so the output at \(a+h\) is less.
For a small negative \(h\), the input has moved to the left of \(a\), and dividing by \(h\) reverses the sign of the comparison. Thus, when \(f'(a)>0\), nearby outputs to the left are lower than \(f(a)\); when \(f'(a)<0\), nearby outputs to the left are higher. These statements compare nearby values with the value at \(a\). They do not automatically compare every pair of inputs in a whole interval.
Why the Sign Gives the Claimed Comparison
Here is a direct justification for the positive case. Let \(L=f'(a)>0\). Since the difference quotient tends to \(L\), it is eventually within \(L/2\) of \(L\). In particular, when \(h\ne 0\) is sufficiently small,
This inequality implies that the quotient is greater than \(L/2\), so it is positive. For \(h>0\), multiplying by \(h\) gives \(f(a+h)-f(a)>0\). For \(h<0\), multiplying by \(h\) gives \(f(a+h)-f(a)<0\). Therefore nearby outputs are above the value at \(a\) on its right and below it on its left.
If instead \(L=f'(a)<0\), take a sufficiently small change so that the quotient differs from \(L\) by less than \(|L|/2\). The quotient is then still negative. Multiplying by positive \(h\) gives a negative output change; multiplying by negative \(h\) gives a positive output change. This proves the corresponding local comparisons for a negative derivative.
Worked Example: Position Decreasing Near a Given Time
Worked Example: A Position Value with a Negative Derivative
Suppose an object’s position, measured in metres, is modelled near \(t=2\) seconds by \(s(t)=20+3t-t^2\). Find \(s'(2)\) from the difference quotient and interpret its sign.
First calculate the position at the specified time:
For a nonzero change \(h\), expand the position at \(2+h\):
Now form and simplify the difference quotient:
Taking the limit as \(h\to0\) gives \(s'(2)=-1\text{ m/s}\). The exact change in position from the value at \(t=2\) is \(s(2+h)-s(2)=-h-h^2\). For \(0<h<1\), this change is negative, so the position value is lower just after \(t=2\) than it is at \(t=2\). For \(-1<h<0\), the change is positive, so the nearby position value to the left is higher.
Check: The simplified quotient is \(-1-h\), whose limit is \(-1\), agreeing with the derivative stated above. Its units are metres per second because position is measured in metres and time in seconds.
Worked Example: Temperature Increasing Near a Time
Worked Example: A Temperature with a Positive Derivative
Let a room’s temperature in degrees Celsius be \(T(t)=10+2t+t^2\), where \(t\) is time in hours. Find the derivative at \(t=2\), and interpret it.
The temperature at that time is
For a nonzero change \(h\), calculate the new temperature and its difference from \(T(2)\):
The difference quotient is therefore
Its limit as \(h\to0\) is \(T'(2)=6^\circ\text{C}/\text{hour}\). For \(0<h<1\), the temperature change is \(h(6+h)>0\). Thus, for a small increase in time after \(t=2\), the temperature is higher than \(18^\circ\text{C}\). For \(-1<h<0\), \(h<0\) and \(6+h>0\), so \(h(6+h)<0\): just before \(t=2\), the temperature is lower than \(18^\circ\text{C}\).
Check: The quotient \(6+h\) tends to \(6\), so the derivative is positive as reported. Its units, degrees Celsius per hour, match temperature change divided by time change.
Worked Example: Cost Decreasing Near a Time
Worked Example: An Operating Cost with a Negative Derivative
Suppose an operating cost in dollars is modelled by \(C(t)=500-12t+t^2\), where \(t\) is time in months. Find the derivative at \(t=4\) and explain what the sign means.
At \(t=4\), the model gives
Expanding at \(4+h\) gives
Consequently, the difference quotient and its limit are
For \(0<h<1\), the cost change is \(h(h-4)<0\). Thus, just after month \(4\), the model gives a lower operating cost than the \(468\) dollars at month \(4\). The negative derivative describes a local decrease in cost as time advances near that input.
Check: The quotient tends to \(-4\), and the exact change \(h(h-4)\) is negative for the stated positive changes. The derivative’s units are dollars per month, as required.
A Pointwise Sign Is Not an Interval Claim
A frequent mistake is to take a derivative sign at one input and extend it to a whole interval. For example, consider \(f(x)=x^3-3x\). Its derivative is \(f'(x)=3x^2-3\), so \(f'(2)=3(2^2)-3=9>0\). It would be wrong to conclude from this single value that \(f\) increases throughout \([0,2]\). In fact, \(f(0)=0\) and \(f(1)=1-3=-2\), so the output falls between these two inputs.
The correction is to keep the conclusion tied to its evidence. A strictly positive derivative at \(x=2\) guarantees a local comparison with \(f(2)\) for sufficiently small changes from \(2\). It does not certify the behaviour across \([0,2]\). To describe a broader interval, information about the derivative across that interval is needed.
For example, the input might be time and the output might be temperature.
A positive value indicates a local rise to the right; a negative value indicates a local fall to the right.
Compare nearby output values with the output at the specified input, and report output units per input unit.
Check Your Understanding
Answer using local comparisons with the value at the specified input. Do not infer behaviour over a whole interval from a derivative at one point.
- If \(f'(a)>0\), what can you say about \(f(a+h)\) compared with \(f(a)\) for sufficiently small positive \(h\)?
- If \(g'(b)<0\), what does the sign say about the output just after input \(b\)?
- A temperature is measured in degrees Celsius and time in minutes. What units should its derivative have?
- Why does a positive derivative at one input not prove that the function increases throughout a specified interval?
- If a cost derivative is negative at a particular time, what local cost comparison does that support for small positive changes in time?
