From a Base Point to a Nearby Value
A function of three variables can change when any one of its inputs changes. Near a point where the function is differentiable, these three changes combine to give a first-order approximation: an affine function that matches the original function’s value and first-order behavior at that point.
The procedure is the same whether the function describes a geometric quantity, a physical model, or an algebraic expression. Choose a base point, compute the function value and its three partial derivatives there, and use those quantities to estimate the value at a nearby point.
Definition and Conditions
Let \(f\) be defined on an open set containing the point \((a,b,c)\). If \(f\) is differentiable at \((a,b,c)\), its linearization there is the affine function \(L\) given below. The quantities \(f_x(a,b,c)\), \(f_y(a,b,c)\), and \(f_z(a,b,c)\) are the three first partial derivatives evaluated at the base point.
The base point must lie within the function’s domain, and differentiability at that point is the condition that justifies this first-order approximation. In common examples, the functions are built from products, roots, and reciprocal powers and are differentiable wherever their formulas are defined with the necessary denominators nonzero and root expressions in their valid range.
Having three partial derivatives at a point does not, by itself, guarantee differentiability there. The earlier material on differentiability establishes this distinction. In particular, the formula’s coefficients can be calculated from partial derivatives, but using the resulting expression as a valid first-order approximation requires differentiability.
Why the Formula Works
Write a nearby point in terms of its coordinate changes: \(x=a+h\), \(y=b+k\), and \(z=c+\ell\). Differentiability at \((a,b,c)\) means that the change in \(f\) is given by its linear part plus a remainder that is negligible compared with the length of the change vector.
where the remainder satisfies the following condition as \((h,k,\ell)\) approaches \((0,0,0)\):
To see why the linearization follows, substitute \(h=x-a\), \(k=y-b\), and \(\ell=z-c\) into the differentiability equation. Its first four terms are exactly \(L(x,y,z)\). Therefore \(f(x,y,z)=L(x,y,z)+r(x-a,y-b,z-c)\). As the point approaches the base point, the remainder becomes negligible relative to the size of the change. This is the precise reason the affine expression provides a first-order approximation nearby.
The formula also gives immediate checks. At the base point, all three coordinate changes are zero, so \(L(a,b,c)=f(a,b,c)\). The coefficients of \(x-a\), \(y-b\), and \(z-c\) are respectively the partial derivatives in the three coordinate directions at that point.
A Practical Construction
Use a point \((a,b,c)\) near the input where the function and its partial derivatives are straightforward to evaluate.
Find \(f(a,b,c)\), \(f_x(a,b,c)\), \(f_y(a,b,c)\), and \(f_z(a,b,c)\).
Multiply each partial derivative by its matching coordinate change, then add the base value.
Use the resulting value of \(L(x,y,z)\) as the first-order estimate of \(f(x,y,z)\).
Keep the variables in the same order throughout. The coefficient of \(x-a\) must come from \(f_x\), the coefficient of \(y-b\) from \(f_y\), and the coefficient of \(z-c\) from \(f_z\). A correct derivative attached to the wrong coordinate change produces a different, generally incorrect, approximation.
Worked Examples
Worked Example: A Product of Three Variables
Estimate \(f(2.02,2.97,4.01)\) for \(f(x,y,z)=xyz\), using the base point \((2,3,4)\).
First evaluate the function and partial derivatives at the base point:
Thus the three partial derivatives at \((2,3,4)\) are
The coordinate changes from the base point to the requested point are \(2.02-2=0.02\), \(2.97-3=-0.03\), and \(4.01-4=0.01\). Therefore,
The linearization estimates the requested value as \(24.06\). Check the construction at the base point: \(L(2,3,4)=24\), matching \(f(2,3,4)\), and its three coefficients \(12,8,6\) match the partial derivatives there. Direct substitution gives \(2.02\cdot2.97=5.9994\) and \(5.9994\cdot4.01=24.057594\), consistent with the nearby estimate.
Worked Example: A Square Root
Estimate \(f(1.04,1.97,6.08)\) for \(f(x,y,z)=\sqrt{x+y+z}\), using the base point \((1,2,6)\). The sum inside the root is positive near this point, so the function and its partial derivatives are defined there.
At the base point, \(x+y+z=1+2+6=9\), and therefore \(f(1,2,6)=\sqrt{9}=3\). Differentiating with respect to each variable gives
Since the denominator at \((1,2,6)\) is \(2\sqrt{9}=6\), each partial derivative there is \(1/6\). Hence
The estimate is \(3.015\). As a check, \(L(1,2,6)=3\), and the coefficient of each coordinate change is the corresponding partial derivative at the base point. Direct substitution into the original function gives \(\sqrt{1.04+1.97+6.08}=\sqrt{9.09}\), which is close to \(3.015\).
Worked Example: A Reciprocal Product
Estimate \(f(2.02,0.99,1.01)\) for \(f(x,y,z)=1/(xyz)\), using the base point \((2,1,1)\). The function is defined near this point because none of the three factors in the denominator is zero.
At the base point, \(f(2,1,1)=1/(2\cdot1\cdot1)=1/2\). Rewriting the function as \((xyz)^{-1}\) and differentiating one variable at a time gives
Evaluation at \((2,1,1)\) yields
The linearization and its value at the target are
Thus the estimate is \(0.495\). At the base point, the linearization gives \(L(2,1,1)=1/2\), as required, and its three coordinate coefficients match the computed partial derivatives. For a direct comparison, the target denominator is \(2.02\cdot0.99\cdot1.01=1.9998\cdot1.01=2.019798\), so the original function is \(1/2.019798\), approximately \(0.495099\). The estimate is close to this value.
Interpreting the Coordinate Contributions
Each term in the linearization has a specific role. The value \(f(a,b,c)\) gives the starting value. The term \(f_x(a,b,c)(x-a)\) estimates the change associated with moving in the \(x\)-coordinate direction; the other two terms do the same for \(y\) and \(z\). Adding these first-order contributions gives the change predicted by the linear model.
This interpretation explains why the matching of coefficients matters. For example, the partial derivative \(f_y(a,b,c)\) measures change in the \(y\)-direction while the other coordinates are held fixed. Its coefficient must therefore multiply \(y-b\), not \(x-a\) or \(z-c\).
A Common Mistake
A frequent error is to compute the correct three partial derivatives but pair them with the wrong coordinate changes. For \(f(x,y,z)=xyz\) at \((2,3,4)\), the correct derivative values are \(12,8,6\), in the order \(x,y,z\). At \((2.02,2.97,4.01)\), swapping the first two changes would give the incorrect calculation
This is not the linearization because \(f_x=12\) must multiply \(x-2=0.02\), and \(f_y=8\) must multiply \(y-3=-0.03\). The corrected calculation is
The simple check is to write each coordinate change directly beside its matching partial derivative before evaluating the expression.
Using Linearizations Responsibly
A linearization is most useful when the target point is near the base point and the function is differentiable there. The three examples show how the same formula applies to different kinds of expressions: a product, a root, and a reciprocal product. In each case, choosing a base point with convenient function and derivative values makes the arithmetic easier.
The result is a first-order estimate, not an identity at every nearby point. Differentiability explains why the remainder is small relative to the size of the input change, but the linearization formula alone does not provide a numerical bound on the approximation’s error. For a careful calculation, keep the exact base-point values and coordinate changes until the final arithmetic step, and check that the linearization returns the correct function value when evaluated at its base point.
Check Your Understanding
Use the definition and the examples above to answer the following questions.
- Write the general formula for the linearization of a differentiable function \(f(x,y,z)\) at \((a,b,c)\).
- For \(f(x,y,z)=x+y+z\), find the linearization at \((1,2,3)\). What does it predict at \((1.1,1.9,3.2)\)?
- Why does the existence of three partial derivatives at a point not, by itself, justify using the linearization as a first-order approximation?
- For \(f(x,y,z)=1/(xyz)\), which partial derivative multiplies \(z-c\) in the linearization at \((a,b,c)\)?
- State two checks that can help detect an incorrectly constructed linearization.