Computing Linearizations of Two-Variable Functions

What a Linearization Does

A function of two variables can be difficult to evaluate exactly, even when the input is close to a point where the function and its derivatives are easy to calculate. If the function is differentiable at that point, its changes nearby are described to first order by a linear function. This linear function is called the linearization.

To construct it, we need three pieces of information at a chosen base point: the function value, the partial derivative with respect to $x$, and the partial derivative with respect to $y$. The base point is written $(a,b)$, and the resulting linear function is written $L(x,y)$.

Definition: Suppose $f$ is differentiable at an interior point $(a,b)$ of its domain. The linearization of $f$ at $(a,b)$ is $$ L(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b). $$

The terms $x-a$ and $y-b$ are the changes in the coordinates from the base point. The coefficients $f_x(a,b)$ and $f_y(a,b)$ describe how the function changes, to first order, in the two coordinate directions. The result is a plane-valued function of the input $(x,y)$.

Why This Formula Works

Differentiability at $(a,b)$ means that when the input changes by small amounts $h$ and $k$, the change in the function is given by a linear part plus a remainder that is negligible compared with the distance moved. More precisely, there is a remainder $r(h,k)$ such that

f(a+h,b+k) = f(a,b)+f_x(a,b)h+f_y(a,b)k+r(h,k), \qquad \frac{r(h,k)}{\sqrt{h^2+k^2}}\longrightarrow 0

as $(h,k)\to(0,0)$, with $(h,k)\ne(0,0)$. Set $h=x-a$ and $k=y-b$. Then $a+h=x$ and $b+k=y$, so the expansion becomes

$$ f(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)+r(x-a,y-b). $$

The expression before the remainder is exactly $L(x,y)$. Thus $f(x,y)$ equals its linearization plus a remainder that becomes negligible compared with the distance from $(x,y)$ to $(a,b)$. This is why $L(x,y)$ provides a first-order approximation near the base point.

The formula is useful only as a justified local approximation when $f$ is differentiable at the chosen point. Having partial derivatives there, by itself, does not guarantee differentiability. The construction of the formula can still be performed from partial derivative values, but without differentiability it need not approximate the function in the stated first-order sense.

A Reliable Construction Process

Choose the base point.
Write down $(a,b)$ and check that the function is defined in a neighborhood of this point.

Find the function value.
Calculate $f(a,b)$, which sets the value of the linearization at the base point.

Find and evaluate the partial derivatives.
Compute $f_x$ and $f_y$, then substitute $(a,b)$ into each derivative.

Insert the coordinate shifts.
Use $x-a$ with $f_x(a,b)$, and $y-b$ with $f_y(a,b)$.

Evaluate at the desired input.
Substitute the nearby point into $L$ to obtain the linear approximation.

Worked Examples

Worked Example: A Polynomial Function

Find the linearization of $f(x,y)=x^2+xy+2y^2$ at $(1,-1)$, then use it to approximate $f(1.1,-0.9)$.

First evaluate the function at the base point:

$$ f(1,-1)=1^2+(1)(-1)+2(-1)^2=1-1+2=2. $$

Differentiate with respect to $x$, holding $y$ fixed, and with respect to $y$, holding $x$ fixed:

f_x(x,y)=2x+y, \qquad f_y(x,y)=x+4y.

Evaluate both partial derivatives at $(1,-1)$:

f_x(1,-1)=2(1)+(-1)=1, \qquad f_y(1,-1)=1+4(-1)=-3.

Substitute these values and the base point into the definition:

\begin{aligned} L(x,y) &=2+1(x-1)-3\bigl(y-(-1)\bigr)\\ &=2+(x-1)-3(y+1)\\ &=x-3y-2. \end{aligned}

At the requested input, the linearization gives

$$ L(1.1,-0.9)=1.1-3(-0.9)-2=1.1+2.7-2=1.8. $$

Therefore, $f(1.1,-0.9)$ is approximated by $1.8$. Check the construction at the base point: $L(1,-1)=1-3(-1)-2=2=f(1,-1)$. Also $L_x=1=f_x(1,-1)$ and $L_y=-3=f_y(1,-1)$, so the linearization has the required value and slopes there.

Worked Example: A Radical Function

Find the linearization of $f(x,y)=\sqrt{9+x+2y}$ at $(0,0)$, then use it to approximate $f(0.1,0.1)$.

The radicand at the base point is $9$, so the function is defined and differentiable in a neighborhood of $(0,0)$ where the radicand remains positive. Its value at the base point is

f(0,0)=\sqrt{9}=3.

Using the chain rule for the square root gives

f_x(x,y)=\frac{1}{2\sqrt{9+x+2y}}, \qquad f_y(x,y)=\frac{2}{2\sqrt{9+x+2y}} =\frac{1}{\sqrt{9+x+2y}}.

At the base point these are

f_x(0,0)=\frac{1}{2\sqrt{9}}=\frac{1}{6}, \qquad f_y(0,0)=\frac{1}{\sqrt{9}}=\frac{1}{3}.

The linearization is therefore

L(x,y)=3+\frac{x}{6}+\frac{y}{3}.

At $(0.1,0.1)$, the linearized value is

L(0.1,0.1) =3+\frac{0.1}{6}+\frac{0.1}{3} =3+\frac{1}{60}+\frac{1}{30} =3+\frac{1}{20} =3.05.

Thus the linearization approximates $\sqrt{9+0.1+2(0.1)}=\sqrt{9.3}$ by $3.05$. As a check, $L(0,0)=3=f(0,0)$, and the coefficients of $x$ and $y$ are respectively the partial derivative values $1/6$ and $1/3$ at the base point.

Worked Example: An Exponential Function

Find the linearization of $f(x,y)=e^{x+2y}$ at $(0,0)$, then approximate $f(0.02,0.01)$.

At the base point,

$$ f(0,0)=e^0=1. $$

By the chain rule, differentiating the exponent with respect to each variable gives

f_x(x,y)=e^{x+2y}, \qquad f_y(x,y)=2e^{x+2y}.

Therefore,

f_x(0,0)=1, \qquad f_y(0,0)=2.

Putting these values into the formula yields

$$ L(x,y)=1+x+2y. $$

At $(0.02,0.01)$, the estimate is

$$ L(0.02,0.01)=1+0.02+2(0.01)=1+0.02+0.02=1.04. $$

So $e^{0.02+2(0.01)}=e^{0.04}$ is approximated by $1.04$. At the base point, $L(0,0)=1=f(0,0)$; its partial derivatives are $L_x=1$ and $L_y=2$, matching the partial derivatives of $f$ there.

A Common Mistake: Omitting the Coordinate Shifts

A frequent error is to use $x$ and $y$ in place of the shifts $x-a$ and $y-b$. For the polynomial example, the incorrect calculation would be

L_{\text{wrong}}(x,y)=2+1x-3y.

This cannot be the correct linearization at $(1,-1)$, because its value there is

L_{\text{wrong}}(1,-1)=2+1+3=6, \qquad f(1,-1)=2.

The linearization must agree with the function at its base point. The correction is to subtract the base coordinates in the corresponding terms:

L(x,y)=2+1(x-1)-3\bigl(y-(-1)\bigr)=x-3y-2.

Now $L(1,-1)=2$, as required. A quick check at the base point is an effective way to catch a missing shift or a sign error.

Using the Linearization Carefully

The formula can be viewed as a calculation recipe, but its terms also have a clear interpretation. The constant $f(a,b)$ anchors the approximation at the base point. The $x$-term accounts for a change in the first coordinate, and the $y$-term accounts for a change in the second. Each derivative must be evaluated at the same base point used in the coordinate shifts.

Linearizations are most useful for inputs near $(a,b)$. Differentiability says the remainder is negligible relative to the distance from the base point as that distance tends to zero; it does not say that the approximation is exact at other points, or provide a numerical error bound for an arbitrary input. For a given problem, choose the base point with care, verify differentiability there, and keep the distinction between the function $f$ and its linear approximation $L$.

Key takeaway: To linearize $f$ at $(a,b)$, evaluate $f$, $f_x$, and $f_y$ at that point, then pair the partial derivatives with $x-a$ and $y-b$. Differentiability is what justifies using the resulting plane as a first-order local approximation.

Check Your Understanding

Work through these questions by identifying the base point, the function value, and both partial derivative values before writing a linearization.

For $f(x,y)=x^2+3y$, what is the general formula for the linearization at $(a,b)$?
For $f(x,y)=xy$, find the linearization at $(2,1)$, and check its value at the base point.
For $f(x,y)=\sqrt{4+x+y}$, find the linearization at $(0,0)$. What condition on the radicand ensures the function is defined near the base point?
For $f(x,y)=e^{2x-y}$, find the linearization at $(0,0)$ and use it to approximate $f(0.01,0.02)$.
Why do the coordinate terms in the formula use $x-a$ and $y-b$, rather than just $x$ and $y$?