From Input Changes to a Linear Prediction
A function of two variables can change when either input changes, or when both change together. At a fixed point, its partial derivatives describe the rates of change in the two coordinate directions. The total differential combines those two rates into one linear expression for the first-order part of the function’s change.
The inputs’ changes will be denoted by \(dx\) and \(dy\). In this setting, these symbols mean actual increments from a chosen base point. The differential \(df\) is then defined using those increments and the partial derivatives at that point. It is important to keep the meaning of \(df\) precise: for a general differentiable function, it is the linear part of the function change, not necessarily the exact change itself.
Input Increments and Function Change
Let \(f\) be a real-valued function of two variables, and choose a base point \((a,b)\) in its domain. If the input moves to \((a+dx,b+dy)\), then \(dx\) and \(dy\) are the changes in the first and second coordinates:
The actual change in the function value is the new value minus the old value. It is often written \(\Delta f\):
This definition gives the exact change, whenever both input points are in the domain. To define the total differential, we also require \(f\) to be differentiable at \((a,b)\). Differentiability means that near this point the function’s change consists of a linear part plus a remainder that becomes negligible relative to the size of the input change.
Here \(f_x(a,b)\) and \(f_y(a,b)\) are the partial derivatives evaluated at the base point, not at the new point. In the common notation \(z=f(x,y)\), one may write \(dz\) in place of \(df\):
In this version, the formula is understood at a particular base point \((x,y)\), with the partial derivatives evaluated there. The symbols \(dx\) and \(dy\) specify the input increments being considered.
Why the Formula Works
The formula is a direct expression of differentiability. At \((a,b)\), differentiability says that the function can be written as a linear approximation plus a remainder:
where the remainder \(r\) satisfies
for nonzero increments. Subtracting \(f(a,b)\) from both sides yields the actual change:
The first two terms are exactly \(df\), so \(\Delta f=df+r(dx,dy)\). This establishes both the formula and its role: \(df\) is the linear part of the change, while the remainder accounts for what that linear part does not capture.
Write the function change as its linear part plus a remainder that is small relative to the input displacement.
The coefficients of \(dx\) and \(dy\) are the partial derivatives at the base point.
By definition, \(df=f_x(a,b)\,dx+f_y(a,b)\,dy\); the actual change also includes the remainder.
The expression is linear in the increments: doubling both increments doubles \(df\), and adding two pairs of increments adds their predicted differentials. That linearity is what makes \(df\) the derivative’s contribution to the function change.
Worked Example: A Quadratic Function with a Linear Term
Worked Example: Computing a Differential and Comparing Changes
Let \(f(x,y)=x^2+3y\), with base point \((2,1)\), and choose \(dx=\frac{1}{10}\), \(dy=-\frac{1}{5}\). Find \(df\), then compare it with the actual function change.
First compute the partial derivatives:
At the base point, these are \(f_x(2,1)=4\) and \(f_y(2,1)=3\). Therefore,
The new input point is
The old function value is \(f(2,1)=2^2+3(1)=7\). The new value is
Thus the actual change is
The differential is \(-\frac{1}{5}=-\frac{20}{100}\), whereas the exact change is \(-\frac{19}{100}\). Substitution back into the change formula gives the remainder:
The calculation checks: the differential is the linear part, and the remaining \(\frac{1}{100}\) makes the total equal the exact change.
Worked Example: A Product and a Mixed Increment
Worked Example: The Differential of a Product Function
Let \(f(x,y)=xy\), with base point \((1,2)\), and let \(dx=\frac{1}{4}\), \(dy=-\frac{1}{2}\). Compute the total differential and the actual change.
The partial derivatives are \(f_x(x,y)=y\) and \(f_y(x,y)=x\). At \((1,2)\), they have values \(2\) and \(1\). Thus,
The new point is \(\left(\frac{5}{4},\frac{3}{2}\right)\). Since the old function value is \(f(1,2)=2\), the exact change is
The values do not agree: \(df=0\), while \(\Delta f=-\frac{1}{8}\). The difference can also be verified directly by expanding the product:
At \(a=1\), \(b=2\), the first two terms give \(df\), and the last term is
Therefore \(\Delta f=df+dx\,dy=0-\frac{1}{8}=-\frac{1}{8}\), agreeing with the direct substitution.
Worked Example: A Function with Two Squared Inputs
Worked Example: Adding the Contributions from Both Variables
Let \(f(x,y)=x^2+y^2\), with base point \((3,4)\), and choose \(dx=-\frac{1}{5}\), \(dy=\frac{1}{10}\). Find the differential and verify the exact change.
The partial derivatives are \(f_x=2x\) and \(f_y=2y\). Evaluating them at the base point gives \(f_x(3,4)=6\) and \(f_y(3,4)=8\). Hence,
The new point is \(\left(3-\frac{1}{5},4+\frac{1}{10}\right)=\left(\frac{14}{5},\frac{41}{10}\right)\). At the old point, \(f(3,4)=9+16=25\). At the new point,
So the actual change is
Since \(df=-\frac{2}{5}=-\frac{8}{20}\), the remainder is
The direct evaluation confirms that the linear differential gives \(-\frac{8}{20}\), and the additional \(\frac{1}{20}\) yields the actual change \(-\frac{7}{20}\).
A Common Mistake: Treating the Differential as an Exact Change
A frequent error is to write \(\Delta f=df\) for every input change. The product example shows why that is not generally valid. For \(f(x,y)=xy\) at \((1,2)\) with \(dx=\frac{1}{4}\) and \(dy=-\frac{1}{2}\), the incorrect calculation would be:
But direct substitution gives \(\Delta f=-\frac{1}{8}\), not \(0\). The corrected relationship is
where the remainder is not necessarily zero. In this product example, the remainder is \(dx\,dy=-\frac{1}{8}\). The differential and the actual change agree in special cases, such as when the function is linear, but not for every differentiable function or every finite input increment.
Why the Total Differential Matters
The total differential gives a compact way to express how the two coordinate directions contribute to the linear part of a function’s change. Each partial derivative supplies a coefficient: the \(x\)-direction contribution is \(f_x(a,b)\,dx\), and the \(y\)-direction contribution is \(f_y(a,b)\,dy\). Adding them gives the derivative’s action on the pair of input increments.
The distinction between a linear prediction and an exact change matters whenever \(dx\) and \(dy\) are not zero. The definition of differentiability controls the remainder as the increments approach zero; it does not assert that the remainder vanishes for each chosen pair. Keeping \(\Delta f\) and \(df\) separate makes the meaning of the formula clear and prevents an approximation from being mistaken for an identity.
Check Your Understanding
Use the definitions and examples above to answer these questions.
- At a base point \((a,b)\), how are \(dx\) and \(dy\) related to the new input point?
- Write the formula for \(df\) and state where its partial derivatives are evaluated.
- What condition on \(f\) is required for the total differential formula?
- How does the actual change \(\Delta f\) relate to the total differential and the remainder?
- For \(f(x,y)=xy\) at \((1,2)\), with \(dx=\frac{1}{4}\) and \(dy=-\frac{1}{2}\), what is the remainder \(\Delta f-df\)?