What an Implicit Rate Describes
An equation involving \(x\) and \(y\) can describe a curve without giving \(y\) explicitly as a function of \(x\). Even so, near some points, the curve may have a branch on which \(y\) changes smoothly as \(x\) changes. On such a branch, \(\frac{dy}{dx}\) describes the instantaneous rate of change of \(y\) with respect to \(x\).
The value is more informative than a bare number. Its sign tells whether \(y\) is increasing or decreasing as \(x\) increases, and its units tell how much of \(y\) changes per unit of \(x\). Near the point where the rate is measured, it can also be used to estimate the effect of a small change in \(x\).
Why the Implicit Slope Is a Rate
Suppose the curve is described by \(F(x,y)=0\), where \(F\) is differentiable, and suppose that near the point of interest the curve can be described by a differentiable function \(y=y(x)\). Along this branch, changing \(x\) also changes \(y\), so the chain rule gives the derivative of the whole expression \(F(x,y(x))\):
Here \(F_x\) and \(F_y\) are the partial derivatives of \(F\) with respect to \(x\) and \(y\). Since every point on the branch satisfies \(F(x,y(x))=0\), its derivative with respect to \(x\) is zero. Therefore,
If \(F_y\ne 0\) at the point, solving this equation gives
This derivation explains why the result is a rate: it is the derivative of \(y\) as \(x\) varies along the curve, not a ratio chosen independently of the curve. The condition \(F_y\ne 0\) is important for this formula and for expressing the branch locally as \(y=y(x)\). If the condition fails, this formula cannot give the rate in that form; the behavior at such a point needs separate analysis.
Units and Signs
The units of \(\frac{dy}{dx}\) are the units of \(y\) divided by the units of \(x\). If \(y\) is measured in meters and \(x\) in seconds, then \(\frac{dy}{dx}\) is measured in meters per second. If both variables use the same units, the units cancel, but the rate still describes how one variable changes relative to the other.
The sign has a direct local interpretation, provided \(x\) is increasing along the branch:
| Rate | Local interpretation as \(x\) increases |
|---|---|
| \(\frac{dy}{dx}>0\) | \(y\) is increasing. |
| \(\frac{dy}{dx}<0\) | \(y\) is decreasing. |
| \(\frac{dy}{dx}=0\) | The instantaneous rate of change of \(y\) is zero. |
A negative rate does not mean that \(y\) itself is negative. It means that \(y\) decreases as \(x\) increases nearby. Likewise, a positive rate does not mean that both variables have positive values; it describes how their values change.
Worked Example: A Circle Measured in Centimeters
Worked Example: A Circle Measured in Centimeters
A point \((x,y)\), measured in centimeters, lies on the circle \(x^2+y^2=25\). At \((3,4)\), interpret \(\frac{dy}{dx}\). Then estimate the change in \(y\) if \(x\) increases by \(0.2\) centimeters.
First check the specified point:
Differentiate the relation along the curve. The derivative of \(x^2\) is \(2x\), and the chain rule gives \(\frac{d}{dx}(y^2)=2y\frac{dy}{dx}\):
At \((3,4)\), substitute \(x=3\) and \(y=4\):
The rate is \(-\frac{3}{4}\) centimeter of \(y\) per centimeter of \(x\). The negative sign says that \(y\) is decreasing as \(x\) increases near \((3,4)\). For an increase \(\Delta x=0.2\) centimeters, the local estimate is
Thus \(y\) is estimated to decrease by \(0.15\) centimeters. The rate and change have the expected units: \((\text{cm}/\text{cm})(\text{cm})=\text{cm}\). As a check on the calculation, the substituted slope satisfies the differentiated equation: \(6+8(-\frac{3}{4})=6-6=0\). The change estimate is local, not an exact finite-change claim.
Worked Example: A Quantity Changing with Time
Worked Example: A Quantity Changing with Time
Suppose \(x\), in hours, and \(y\), in dollars per hour, satisfy \(xy=60\) dollars. At the point \(x=2\) hours and \(y=30\) dollars per hour, find and interpret \(\frac{dy}{dx}\). Estimate the change in \(y\) for an increase of \(0.1\) hour in \(x\).
The point satisfies the relation because
Differentiate \(xy=60\) with respect to \(x\). The product rule gives \(y+x\frac{dy}{dx}=0\), since the derivative of the constant is zero:
Substitute the point and solve:
The units are \((\text{dollars/hour})/\text{hour}\), or dollars per hour squared. The negative sign means that \(y\), the rate measured in dollars per hour, decreases as \(x\) increases near the point. For \(\Delta x=0.1\) hour,
So the local estimate is a decrease of \(1.5\) dollars per hour. The units reduce correctly, and the slope can be checked in the differentiated equation: \(30+2(-15)=0\). This example also shows why keeping track of units matters: the derivative is not simply “\(-15\)”; its units are part of its meaning.
Worked Example: Reading a Rate in Distance Units
Worked Example: Reading a Rate in Distance Units
A curve is given by \(x^2+xy+y^2=7\text{ km}^2\), with \(x\) and \(y\) measured in kilometers. Find the rate at \((1\text{ km},2\text{ km})\), and estimate the change in \(y\) if \(x\) increases by \(0.05\) kilometer.
Check that the point is on the curve:
Differentiate each term along the curve. The product rule gives \(\frac{d}{dx}(xy)=y+x\frac{dy}{dx}\):
Collect the terms containing \(\frac{dy}{dx}\):
At \((1,2)\), this becomes
Because both variables are measured in kilometers, the slope has units of kilometers per kilometer. It is negative, so \(y\) decreases as \(x\) increases nearby. For \(\Delta x=0.05\) kilometer,
The estimate is a decrease of \(40\) meters. The substituted slope also makes the differentiated equation equal zero: \(4+5(-\frac{4}{5})=4-4=0\). The distance conversion is exact because \(0.04\text{ km}=40\text{ m}\).
A Common Sign Mistake
When differentiating \(x^2+y^2=25\), a common error is to lose the negative sign while solving for the rate:
The first implication is incorrect. Subtracting \(2x\) from both sides gives \(2y\frac{dy}{dx}=-2x\), not \(2x\). At \((3,4)\), the incorrect calculation would give \(+\frac{3}{4}\), while the correct calculation gives \(-\frac{3}{4}\). The correct sign matches the local behavior: near the upper-right part of the circle, increasing \(x\) corresponds to decreasing \(y\).
Check Your Understanding
For each question, distinguish the rate at a point from a finite change in the variables.
- If \(\frac{dy}{dx}=-2\) at a point and \(x\) increases slightly, what does the sign say about \(y\)?
- If \(x\) is measured in seconds and \(y\) in meters, what are the units of \(\frac{dy}{dx}\)?
- For \(x^2+y^2=13\), find \(\frac{dy}{dx}\) at \((2,3)\), and state what the sign means locally.
- If \(\frac{dy}{dx}=4\) at a point, estimate \(\Delta y\) for a small increase \(\Delta x=0.05\). Why is the result an estimate rather than necessarily an exact change?
- In the relation \(xy=60\) dollars, where \(x\) is measured in hours and \(y\) in dollars per hour, what are the units of \(\frac{dy}{dx}\)?
