Estimating Function Changes with Differentials

From Input Changes to an Estimated Output Change

A change in either input of a two-variable function can change its output. If the function is differentiable, its partial derivatives at a starting point give a simple way to estimate that output change. The estimate is especially useful when the input changes are small, but it is not generally the exact change.

For a function $f(x,y)$, suppose the starting input is $(a,b)$, and the inputs change by $\Delta x$ and $\Delta y$. The new input is $(a+\Delta x,b+\Delta y)$. The exact change in the function value is the new output minus the starting output. The differential gives a first-order estimate of that difference.

Definition: If $f$ is differentiable at $(a,b)$, its total differential at that point is $$ df=f_x(a,b)\,dx+f_y(a,b)\,dy. $$ When estimating the effect of specified input changes, set $dx=\Delta x$ and $dy=\Delta y$. The differential estimate of the function change is then $\Delta f\approx df$.

The partial derivatives in this formula are evaluated at the starting point $(a,b)$. The exact function change is

\Delta f=f(a+\Delta x,b+\Delta y)-f(a,b).

The symbols have different roles: $\Delta f$ denotes the actual change, while $df$ denotes the linear estimate of that change. In many applications, the estimate is a good practical substitute when the increments are small. It remains an approximation unless the function is linear in the relevant region or a special cancellation makes the two values agree.

Why the Differential Estimates the Change

Differentiability at $(a,b)$ means that the change in $f$ can be written as a linear function of the input increments plus a remainder that becomes small relative to the size of those increments. More precisely, for increments not both zero,

\Delta f = f_x(a,b)\Delta x+f_y(a,b)\Delta y+r(\Delta x,\Delta y),

where the remainder satisfies

\lim_{(\Delta x,\Delta y)\to(0,0)} \frac{r(\Delta x,\Delta y)} {\sqrt{(\Delta x)^2+(\Delta y)^2}}=0.

The first two terms are exactly the total differential after setting $dx=\Delta x$ and $dy=\Delta y$. Thus the difference between the actual change and its differential estimate is the remainder:

\Delta f-df=r(\Delta x,\Delta y).

As the input changes approach zero, this remainder becomes small relative to the length of the input change. That is why the differential provides a local first-order estimate. Differentiability does not say that the remainder is zero for a particular nonzero change, nor does it guarantee a specified level of accuracy for a large change.

A Reliable Calculation Procedure

For a concrete estimate, keep the base point, input increments, and function changes distinct. The calculation follows directly from the definition.

Record the starting point and increments.
Write the base point as $(a,b)$, and record the specified changes $\Delta x$ and $\Delta y$. The new input is $(a+\Delta x,b+\Delta y)$.

Find the partial derivatives.
Compute $f_x$ and $f_y$, then evaluate both at the starting point $(a,b)$.

Calculate the differential estimate.
Substitute $dx=\Delta x$ and $dy=\Delta y$ into $df=f_x(a,b)\,dx+f_y(a,b)\,dy$.

Find the exact change for comparison.
Evaluate $f$ at the new input and subtract its value at the starting point.

Worked Examples

Worked Example: A Polynomial Function

Estimate the change in $f(x,y)=x^2+3xy+y^2$ as the input changes from $(1,2)$ by $\Delta x=0.1$ and $\Delta y=-0.05$. Then compare the estimate with the exact change.

First compute the partial derivatives:

f_x=2x+3y, \qquad f_y=3x+2y.

At the starting point $(1,2)$,

f_x(1,2)=2(1)+3(2)=8, \qquad f_y(1,2)=3(1)+2(2)=7.

Set $dx=0.1$ and $dy=-0.05$. Then

$$ df=8(0.1)+7(-0.05)=0.8-0.35=0.45. $$

The differential estimates an increase of $0.45$. For the exact change, the new point is \[ (1+0.1,2-0.05)=(1.1,1.95). \] The starting output is

$$ f(1,2)=1^2+3(1)(2)+2^2=1+6+4=11. $$

At the new point,

$$ f(1.1,1.95) =(1.1)^2+3(1.1)(1.95)+(1.95)^2 =1.21+6.435+3.8025 =11.4475. $$

Therefore the exact change is $11.4475-11=0.4475$. The differential estimate $0.45$ differs from the exact change by $0.45-0.4475=0.0025$. The comparison confirms that the estimate is close here, but not exact.

Worked Example: A Function Involving a Square Root

Let $f(x,y)=\sqrt{x^2+y^2}$, and estimate its change from $(3,4)$ when $\Delta x=0.06$ and $\Delta y=-0.08$. Compare with the exact change.

The partial derivatives are

f_x(x,y)=\frac{x}{\sqrt{x^2+y^2}}, \qquad f_y(x,y)=\frac{y}{\sqrt{x^2+y^2}},

where the formula applies when $x^2+y^2>0$. At $(3,4)$, the denominator is $\sqrt{3^2+4^2}=5$, so

f_x(3,4)=\frac{3}{5}, \qquad f_y(3,4)=\frac{4}{5}.

The differential estimate is

df=\frac{3}{5}(0.06)+\frac{4}{5}(-0.08) =0.036-0.064 =-0.028.

For the exact change, the new point is $(3.06,3.92)$. Its output is

f(3.06,3.92) =\sqrt{(3.06)^2+(3.92)^2} =\sqrt{9.3636+15.3664} =\sqrt{24.73} \approx 4.97293.

The starting output is $f(3,4)=5$. Thus the exact change is $\sqrt{24.73}-5$, approximately $4.97293-5=-0.02707$, rounded to five decimal places. The estimated change, $-0.028$, is close to the exact change, but it predicts a slightly larger decrease.

Worked Example: An Exponential Function

For $f(x,y)=e^{x-y}$, estimate the change from $(0,0)$ when $\Delta x=0.1$ and $\Delta y=0.2$, then compare with the exact change.

The partial derivatives are

f_x(x,y)=e^{x-y}, \qquad f_y(x,y)=-e^{x-y}.

At $(0,0)$, these values are $f_x(0,0)=e^0=1$ and $f_y(0,0)=-e^0=-1$. Therefore,

$$ df=1(0.1)+(-1)(0.2)=0.1-0.2=-0.1. $$

The differential predicts a decrease of $0.1$. The new input is $(0.1,0.2)$, and the starting output is $f(0,0)=e^0=1$. At the new input,

f(0.1,0.2)=e^{0.1-0.2}=e^{-0.1}\approx 0.904837.

The exact change is $e^{-0.1}-1\approx 0.904837-1=-0.095163$, with the decimal rounded to six places. The differential estimate $-0.1$ is close to, but not identical to, the exact change.

Reading the Estimate Correctly

The sign of $df$ indicates the direction of change predicted by the differential: a positive value predicts an increase, and a negative value predicts a decrease. Its size estimates the amount of that change. Both input increments matter, including their signs. For example, if $\Delta y<0$, then the contribution $f_y(a,b)\Delta y$ has the opposite sign from $f_y(a,b)$ when that partial derivative is positive.

The estimate is a sum of two contributions. The term $f_x(a,b)\Delta x$ describes the first-order effect associated with changing $x$; $f_y(a,b)\Delta y$ describes the first-order effect associated with changing $y$. These are combined at the same base point. This is not a claim that the function's exact change can be separated into two independent effects for arbitrary finite increments; it is the linear approximation supplied by differentiability.

Comparing an estimate with an exact change is useful when the exact calculation is available: it shows the size and direction of the approximation error for that particular input change. It does not, by itself, establish an error bound for other changes. The remainder in the differentiability statement explains why the approximation improves locally, but a particular accuracy guarantee would require additional information.

Common Pitfall: Evaluating at the New Point

A frequent error is to evaluate the partial derivatives at the endpoint rather than at the starting point. In the polynomial example, the endpoint is $(1.1,1.95)$. At that point,

$$ f_x(1.1,1.95)=2(1.1)+3(1.95)=2.2+5.85=8.05, $$ $$ f_y(1.1,1.95)=3(1.1)+2(1.95)=3.3+3.9=7.2. $$

Using these endpoint values would give the incorrect differential calculation

$$ 8.05(0.1)+7.2(-0.05)=0.805-0.36=0.445. $$

The correction is to evaluate both partial derivatives at $(1,2)$, the point from which the change begins. That gives $8(0.1)+7(-0.05)=0.45$, the appropriate first-order estimate at the specified base point. The endpoint derivatives may be used as part of a different approximation method, but they are not the derivatives in the total differential based at $(1,2)$.

Check Your Understanding

For each question, distinguish the estimated change from the exact change where appropriate.

For $f(x,y)=x^2+y^2$ at $(1,2)$, with $\Delta x=0.02$ and $\Delta y=-0.03$, what is the differential estimate $df$?
In the total differential estimate, where are $f_x$ and $f_y$ evaluated?
What expression gives the exact change in $f$ for increments $\Delta x$ and $\Delta y$ from $(a,b)$?
For a differentiable function, what does the remainder condition say about the error relative to the length of the input change as that change approaches zero?
Why does a negative value of $df$ not prove that the exact change equals that value?