Linearization Workshop with Exponentials and Trigonometric Functions

Using First-Order Approximations with Familiar Functions

Exponential, logarithmic, and trigonometric expressions appear in models of growth, measurement, oscillation, and geometry. When the inputs change slightly, a first-order approximation can estimate the resulting change without evaluating the full function at every nearby point.

The essential work is to evaluate the function and each partial derivative at the chosen base point. A formula for a linearization is only useful once those values have been found correctly. The examples below focus on that evaluation, including the domain and angle conditions that matter.

Definition: If $f$ is differentiable near $(a,b)$, its linearization at $(a,b)$ is $$ L(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b). $$ For inputs near $(a,b)$, $L(x,y)$ is the first-order estimate of $f(x,y)$.

Here $f_x(a,b)$ and $f_y(a,b)$ mean that the partial derivatives are evaluated at the base point, not at the new input. The increments $x-a$ and $y-b$ are measured from that same point. For a function of three variables, the corresponding linearization includes one term for each variable and its partial derivative.

Why the Formula Uses the Derivatives at the Base Point

Write a nearby point as $(a+h,b+k)$, where $h$ and $k$ are its changes from the base point. Differentiability means that the function change is given by its derivative’s linear part, together with a remainder that becomes negligible compared with the distance to the base point:

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 $(h,k)$ approaches $(0,0)$, the remainder is small compared with the size of the input change. Dropping that remainder gives the first-order estimate. Replacing $h$ and $k$ with $x-a$ and $y-b$ yields the stated linearization formula. This also explains why the partial derivatives must be evaluated at the base point: they describe the linear part of the function’s change there.

Choose a base point.
Identify $(a,b)$ and confirm that the function is defined there and nearby.

Find the partial derivatives.
Differentiate with respect to each variable, keeping the other variable fixed.

Evaluate at the base point.
Compute $f(a,b)$, $f_x(a,b)$, and $f_y(a,b)$ before substituting into the linearization.

Apply the estimate nearby.
Use the actual input changes $x-a$ and $y-b$, then check that the result agrees with the function value and slopes at the base point.

Worked Exponential Approximation

Worked Example: Estimating an Exponential Near a Nonzero Base Point

Find the linearization of $f(x,y)=e^{x+2y}$ at $(a,b)=(\ln 2,0)$, then use it to estimate $f(\ln 2+0.03,0.02)$.

First evaluate the function at the base point:

f(\ln 2,0)=e^{\ln 2+2(0)}=e^{\ln 2}=2.

Differentiate with respect to each variable. The derivative of the exponent with respect to $x$ is $1$; with respect to $y$ it is $2$. Thus

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

At the base point, these become

f_x(\ln 2,0)=2, \qquad f_y(\ln 2,0)=2(2)=4.

Therefore the linearization is

L(x,y)=2+2(x-\ln 2)+4y.

The input changes are $x-\ln 2=0.03$ and $y-0=0.02$. Substitution gives

L(\ln 2+0.03,0.02) = 2+2(0.03)+4(0.02) = 2+0.06+0.08 = 2.14.

So the first-order estimate is $f(\ln 2+0.03,0.02)\approx 2.14$. The exact value is $e^{\ln 2+0.03+2(0.02)}=2e^{0.07}$, which is approximately $2.145016$ to six decimal places. The estimate is close, as expected for a nearby input.

To check the linearization at the base point, substitute $x=\ln 2$ and $y=0$: $L(\ln 2,0)=2=f(\ln 2,0)$. Its coefficients of $x-\ln 2$ and $y$ are $2$ and $4$, matching $f_x(\ln 2,0)$ and $f_y(\ln 2,0)$.

Worked Logarithmic Approximation

Worked Example: Checking the Logarithm’s Domain Before Estimating

Use a linearization to estimate $f(1.02,0.04)$ for $f(x,y)=\ln(1+x^2+y)$, based at $(1,0)$.

The logarithm requires its argument to be positive. At the base point the argument is $1+1^2+0=2>0$. At the target point it is $1+(1.02)^2+0.04=1+1.0404+0.04=2.0804>0$. The function is defined at both points; in particular, its argument remains positive in a neighborhood of $(1,0)$.

At the base point,

f(1,0)=\ln(1+1^2+0)=\ln 2.

Differentiate with respect to $x$ and $y$:

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

Evaluating at $(1,0)$ gives

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

Hence the linearization is

L(x,y)=\ln 2+(x-1)+\frac{1}{2}y.

At the target, $x-1=1.02-1=0.02$ and $y-0=0.04$. Thus

L(1.02,0.04) = \ln 2+0.02+\frac{1}{2}(0.04) = \ln 2+0.02+0.02 = \ln 2+0.04 \approx 0.733147.

The estimate is $f(1.02,0.04)\approx 0.733147$. Direct evaluation gives $\ln(2.0804)\approx 0.732560$, to six decimal places. The approximation is not exact, but it is close at this input.

For the check, $L(1,0)=\ln 2=f(1,0)$. The coefficients of $x-1$ and $y$ are $1$ and $1/2$, exactly the partial derivatives evaluated at the base point.

Worked Trigonometric Approximation

Worked Example: Combining Sine and Cosine at an Exact Base Point

Approximate $f(x,y)=\sin x\cos y$ near $(a,b)=(\pi/6,\pi/3)$, and estimate the value at $x=\pi/6+0.04$, $y=\pi/3-0.02$. Angles in this example are in radians.

At the base point, $\sin(\pi/6)=1/2$ and $\cos(\pi/3)=1/2$, so

f\left(\frac{\pi}{6},\frac{\pi}{3}\right) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}.

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

f_x(x,y)=\cos x\cos y, \qquad f_y(x,y)=-\sin x\sin y.

Using $\cos(\pi/6)=\sqrt{3}/2$, $\sin(\pi/3)=\sqrt{3}/2$, and the values above,

f_x\left(\frac{\pi}{6},\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\cdot\frac{1}{2} = \frac{\sqrt{3}}{4}, \qquad f_y\left(\frac{\pi}{6},\frac{\pi}{3}\right) = -\frac{1}{2}\cdot\frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{4}.

The resulting linearization is

L(x,y) = \frac{1}{4} +\frac{\sqrt{3}}{4}\left(x-\frac{\pi}{6}\right) -\frac{\sqrt{3}}{4}\left(y-\frac{\pi}{3}\right).

For the chosen input, the increments are $0.04$ and $-0.02$. Therefore

L\left(\frac{\pi}{6}+0.04,\frac{\pi}{3}-0.02\right) = \frac{1}{4} +\frac{\sqrt{3}}{4}(0.04) -\frac{\sqrt{3}}{4}(-0.02) = \frac{1}{4}+0.015\sqrt{3} \approx 0.275981.

The estimate is approximately $0.275981$, rounded to six decimal places. At the base point, the linearization gives $1/4$, the function value. Its two coefficients match the evaluated partial derivatives $\sqrt{3}/4$ and $-\sqrt{3}/4$, confirming the base-point calculation.

A Common Coefficient Error

When differentiating an exponential of an expression, it is easy to lose a factor from the exponent. For $f(x,y)=e^{x+2y}$, an incorrect calculation might be

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

This omits the derivative of $x+2y$ with respect to $y$, which is $2$. The correct calculation is

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

At $(\ln 2,0)$, the incorrect derivative would give $2$ instead of $4$. The resulting incorrect linearization, $L_{\mathrm{wrong}}(x,y)=2+2(x-\ln 2)+2y$, would estimate the earlier input as $2+0.06+0.04=2.10$, rather than $2.14$. To avoid this mistake, differentiate the exponent with respect to the variable first, then multiply by the exponential and evaluate the full derivative at the base point.

What to Check Before Trusting an Estimate

The linearization is local: it is intended for inputs near the base point, and it is not generally an exact formula for the function elsewhere. An estimate can be less accurate when the input changes are larger. For logarithms, also check that the argument is positive at the base point and remains positive for the nearby inputs under consideration. Exponential functions are defined for every real exponent, while sine and cosine accept real inputs; their angle inputs here are understood in radians.

Function type	Useful check
Exponential	Include the derivative of the exponent for each variable.
Logarithmic	Verify that the logarithm’s argument is positive.
Trigonometric	Evaluate the trigonometric factors and their derivatives at the stated base point.

Across all three types, the same final checks apply: the linearization should reproduce the function value at the base point, and its coefficients should equal the partial derivatives there. These checks catch many errors before an estimate is used.

Check Your Understanding

For each question, identify the base-point values and derivatives before forming an estimate.

For $f(x,y)=e^{2x-y}$, find the linearization at $(0,0)$.
For $g(x,y)=\ln(x+y)$, what condition on $x+y$ is required for $g$ to be defined? At $(1,1)$, find $g$, $g_x$, and $g_y$.
Find the linearization of $h(x,y)=\sin x+\cos y$ at $(0,0)$.
For $q(x,y)=e^{x+3y}$, state the factor that must appear in $q_y$, and explain its source.
What two base-point checks can be used to verify a proposed linearization?