Calculus Solver Instantaneous rate of change calculator
The instantaneous rate of change of f at a point a is the derivative f'(a). It is the limit of the average rate of change over [a, a + h] as h shrinks to zero. The calculator below computes it and shows the average rates converging, which is the part that makes the definition make sense.
Calculator
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Instantaneous rate of change
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How the average rate closes in on it
| Interval width h | Average rate on [a, a+h] |
|---|
Average against instantaneous
The average rate of change over an interval is rise over run: (f(b) − f(a)) / (b − a). It answers what happened across a stretch. A car that covers 120 km in two hours averaged 60 km/h, and that says nothing about its speed at any given moment.
The instantaneous rate is the speedometer reading: the rate at one instant. You get it by taking the average over shorter and shorter intervals starting at a, and seeing what number they settle on.
The definition, and why it has a limit in it
f'(a) = lim as h approaches 0 of (f(a + h) − f(a)) / h.
You cannot just set h = 0, because that gives 0/0, which is undefined. That is precisely why the limit is there: it asks where the quotient is heading as h shrinks, without ever requiring h to actually be zero.
In the table above, watch the average rate stop moving as h gets small. That settling value is the limit, and it is the derivative.
Working one by hand
Take f(x) = x² at a = 3. The quotient is ((3 + h)² − 9) / h. Expand the numerator: 9 + 6h + h² − 9 = 6h + h². Divide by h to get 6 + h.
Now the limit is easy, because the problematic h in the denominator is gone. As h approaches 0, 6 + h approaches 6. So f'(3) = 6.
That cancellation is the whole trick of every derivative from the definition: rearrange until the h in the denominator disappears, then substitute.
What the number means
The instantaneous rate of change is the slope of the tangent line at that point, which is why the two phrases get used interchangeably.
A positive value means the function is rising at that point, negative means falling, and zero means it is momentarily flat, which is where maxima, minima and inflection points live.
The units come along for the ride. If f is metres and x is seconds, f'(a) is in metres per second. If f is cost and x is units produced, f'(a) is the marginal cost.
The same idea under its different names
| Phrase | Means |
|---|---|
| Instantaneous rate of change at a | f'(a) |
| Slope of the tangent at a | f'(a) |
| Derivative evaluated at a | f'(a) |
| Average rate of change on [a, b] | (f(b) − f(a)) / (b − a) |
| Slope of the secant through a and b | (f(b) − f(a)) / (b − a) |
Questions
Is the instantaneous rate of change the same as the derivative?
Yes. The derivative at a point is the instantaneous rate of change at that point. The derivative as a function, f'(x), gives you that rate at every point at once.
Why can't I set h to zero straight away?
Because the quotient becomes 0/0, which is undefined rather than equal to anything. Simplify the expression first so h cancels out of the denominator, and only then let h go to zero.
How accurate is a numerical answer like this one?
The calculator uses a central difference with a very small step, which for ordinary smooth functions matches the exact derivative to several decimal places. Where a function has a corner or a jump at the point, no numerical method is meaningful, because the derivative does not exist there.
What if the rate of change comes out as zero?
The tangent is horizontal at that point, so the function is neither rising nor falling for that instant. It is a critical point, and it is a candidate for a maximum, a minimum, or a point of inflection. Which one it is takes a further test.
More calculus reference
Stuck on the actual problem
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