Tutorials › Calculus III › Common Errors in Differentiability and Linearization

Home  /  Tutorials  /  Differentiability and Linearization

Differentiability and Linearization · Tutorial 349 of 999

Common Errors in Differentiability and Linearization

Diagnose common errors in multivariable differentiability and linearization by checking the definition, the base point, mixed terms, and the meaning of a differential.

Intermediate 9 min read

What You'll Learn

  • State the remainder condition that defines differentiability at a point.
  • Explain why existing partial derivatives alone do not establish differentiability.
  • Identify when a mixed term belongs in a second-order approximation.
  • Calculate increments relative to the correct base point.
  • Distinguish a differential estimate from the exact function change.
  • Use continuity of partial derivatives as a sufficient differentiability test.

Why Small Errors in Setup Matter

Linearization and local approximation depend on more than correctly calculating partial derivatives. The point where derivatives are evaluated, the meaning of the increments, and the order of the approximation all matter. A formula can look plausible and still fail because one of these ingredients has been misused.

A particularly important distinction is between having partial derivatives and being differentiable. Partial derivatives measure changes along coordinate directions. Differentiability requires a stronger statement: one linear function must approximate the change in every nearby direction, with an error that becomes negligible compared with the distance moved.

The Definition to Check

Let \(f\) be a function of two variables, and let \((a,b)\) be an interior point of its domain. The increments from this base point are \(h=x-a\) and \(k=y-b\). Differentiability at \((a,b)\) means that the function can be written as its linearization plus a remainder whose size is negligible relative to the distance \(\sqrt{h^2+k^2}\).

Definition: The function \(f\) is differentiable at \((a,b)\) if there are numbers \(A\) and \(B\) such that
$$ f(a+h,b+k) = f(a,b)+Ah+Bk+r(h,k), \qquad \lim_{(h,k)\to(0,0)} \frac{r(h,k)}{\sqrt{h^2+k^2}}=0. $$
When the partial derivatives exist, differentiability requires \(A=f_x(a,b)\) and \(B=f_y(a,b)\). Thus the linearization is
$$ L(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b). $$

The limit condition says that the error is small compared with the size of the input change, not merely that the error tends to zero. For example, an error proportional to \(\sqrt{h^2+k^2}\) does not meet this condition; an error proportional to \(h^2+k^2\) does.

Theorem: If \(f\) is differentiable at \((a,b)\), then \(f\) is continuous there and its partial derivatives exist, with \(A=f_x(a,b)\) and \(B=f_y(a,b)\). The converse is not guaranteed: existing partial derivatives alone do not imply differentiability.

To see why differentiability implies continuity, write the change as \(Ah+Bk+r(h,k)\). As \((h,k)\) tends to \((0,0)\), the linear terms tend to zero, and the remainder does too because it is \(o(\sqrt{h^2+k^2})\). To identify the partial derivatives, set \(k=0\) and divide by \(h\), or set \(h=0\) and divide by \(k\). The remainder divided by the corresponding increment tends to zero, so the two partial derivatives must be \(A\) and \(B\).

Worked Example: Partial Derivatives That Do Not Give Differentiability

Worked Example: Partial Derivatives That Do Not Give Differentiability

Define \(f(x,y)=\frac{xy}{x^2+y^2}\) when \((x,y)\ne(0,0)\), and set \(f(0,0)=0\). Check whether the partial derivatives at the origin establish differentiability.

For the partial derivative with respect to \(x\), hold \(y=0\). For every nonzero \(h\), \(f(h,0)=0\). Therefore

$$ f_x(0,0) = \lim_{h\to0}\frac{f(h,0)-f(0,0)}{h} = \lim_{h\to0}\frac{0-0}{h} =0. $$

For the partial derivative with respect to \(y\), hold \(x=0\). For every nonzero \(k\), \(f(0,k)=0\), so

$$ f_y(0,0) = \lim_{k\to0}\frac{f(0,k)-f(0,0)}{k} = \lim_{k\to0}\frac{0-0}{k} =0. $$

If \(f\) were differentiable at the origin, its linearization would consequently be \(L(x,y)=0\), and the remainder condition would require \(f(h,k)/\sqrt{h^2+k^2}\to0\). Test the path \(h=k=t\), where \(t\ne0\): \(f(t,t)=t^2/(2t^2)=1/2\). Hence

$$ \frac{f(t,t)-L(t,t)}{\sqrt{t^2+t^2}} = \frac{1/2}{\sqrt{2}|t|}. $$

This expression does not tend to zero; it grows without bound as \(t\to0\). Thus \(f\) is not differentiable at \((0,0)\). In fact, the same path shows that \(f\) is not continuous there, since the function values are \(1/2\) along that path while \(f(0,0)=0\).

Check: The coordinate-axis calculations correctly give both partial derivatives as zero, but the diagonal path violates the differentiability remainder condition. The mistake would be to treat those two partial derivatives as proof of differentiability.

A Reliable Sufficient Test

Although existing partial derivatives are not enough by themselves, continuity of the first partial derivatives in a neighborhood supplies a useful sufficient condition. The reason is that small changes in the function can be split into a horizontal change and a vertical change. The one-variable mean value theorem controls each change, and continuity makes the resulting derivative values close to those at the base point.

Theorem: Suppose \(f_x\) and \(f_y\) exist throughout an open neighborhood of \((a,b)\) and are continuous at \((a,b)\). Then \(f\) is differentiable at \((a,b)\), with linearization \(L(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)\).
1
Split the change.
For sufficiently small \(h,k\), write \(f(a+h,b+k)-f(a,b)\) as the change from \((a,b+k)\) to \((a+h,b+k)\), followed by the change from \((a,b)\) to \((a,b+k)\).
2
Apply the one-variable mean value theorem.
For intermediate values \(\xi\) between \(a\) and \(a+h\), and \(\eta\) between \(b\) and \(b+k\), the change is \(f_x(\xi,b+k)h+f_y(a,\eta)k\).
3
Subtract the proposed linear part.
The remainder is \([f_x(\xi,b+k)-f_x(a,b)]h+[f_y(a,\eta)-f_y(a,b)]k\).
4
Use continuity.
As \((h,k)\to(0,0)\), both bracketed differences tend to zero. Also, \(|h|/\sqrt{h^2+k^2}\le1\) and \(|k|/\sqrt{h^2+k^2}\le1\). Dividing the remainder by \(\sqrt{h^2+k^2}\) therefore gives a quantity tending to zero, as required.

This is a sufficient test, not a necessary one: differentiability can hold even if a first partial derivative is not continuous at the point. Use the test in the direction it guarantees, rather than treating continuity as part of the definition.

Worked Example: Keeping the Mixed Term

Worked Example: Keeping the Mixed Term

Consider \(f(x,y)=xy\) near \((0,0)\). The linearization there is zero because \(f(0,0)=0\), \(f_x(0,0)=0\), and \(f_y(0,0)=0\). The question is whether a second-order approximation retains a mixed term.

The second partial derivatives are \(f_{xx}=0\), \(f_{yy}=0\), and \(f_{xy}=f_{yx}=1\). The second-order term at increments \(h,k\) is therefore

$$ \frac12 \begin{bmatrix}h&k\end{bmatrix} \begin{bmatrix}0&1\\1&0\end{bmatrix} \begin{bmatrix}h\\k\end{bmatrix} = \frac12(hk+kh) = hk. $$

Thus the second-order polynomial is \(P_2(h,k)=hk\), and here it equals the function exactly: \(f(h,k)=hk\). For \(h=0.1\) and \(k=0.2\), the mixed term is \(hk=(0.1)(0.2)=0.02\), matching \(f(0.1,0.2)=0.02\). If the mixed term were omitted, the approximation would be zero, leaving an error of \(0.02\).

There is an important distinction: the zero linearization is still a valid first-order approximation. Its error is \(hk\), and

$$ \frac{|hk|}{\sqrt{h^2+k^2}} \le \frac{(h^2+k^2)/2}{\sqrt{h^2+k^2}} = \frac{\sqrt{h^2+k^2}}{2} \longrightarrow 0. $$

So omitting \(hk\) is not an error when writing the first-order linearization; it is an error when claiming to include all second-order terms. The approximation order determines which terms belong.

Worked Example: Measuring Increments from the Base Point

Worked Example: Measuring Increments from the Base Point

Use the linearization of \(f(x,y)=x^2+y^2\) at \((1,2)\) to estimate the change when the input moves to \((1.1,1.8)\). Compare the estimate with the exact change.

At the base point, \(f(1,2)=1^2+2^2=5\). The partial derivatives are \(f_x=2x\) and \(f_y=2y\), so \(f_x(1,2)=2\) and \(f_y(1,2)=4\). The increments are measured from \((1,2)\): \(h=1.1-1=0.1\) and \(k=1.8-2=-0.2\). Thus

$$ L(1.1,1.8) = 5+(2)(0.1)+(4)(-0.2) = 5+0.2-0.8 = 4.4. $$

The linearization estimates the change as \(4.4-5=-0.6\). Direct calculation gives \(f(1.1,1.8)=1.1^2+1.8^2=1.21+3.24=4.45\), so the exact change is \(4.45-5=-0.55\). The approximation error in the function value is \(4.45-4.4=0.05\).

A common wrong calculation uses the target coordinates as increments, taking \(h=1.1\) and \(k=1.8\). This gives \(5+(2)(1.1)+(4)(1.8)=5+2.2+7.2=14.4\), which is not a local estimate at \((1,2)\). The correction is to subtract the base point first. Substitution into the linearization confirms the correct estimate \(4.4\); direct evaluation confirms the exact value \(4.45\).

A Differential Is an Estimate, Not an Exact Change

At a fixed base point \((a,b)\), the total differential is the linear expression \(df=f_x(a,b)\,dx+f_y(a,b)\,dy\). When \(dx=h\) and \(dy=k\), it gives the linearized change. The exact change is \(f(a+h,b+k)-f(a,b)\), which may include higher-order contributions. The two agree only when the function and increments make that exact; in general, the differential is an approximation.

The preceding example makes the distinction concrete: \(df=-0.6\), while the exact change is \(-0.55\). Substituting the computed values back into the definitions verifies both quantities. Confusing them can turn a useful estimate into an incorrect claim of equality.

Common checks:
  • To claim differentiability, verify the remainder condition or use a valid sufficient test.
  • For a Taylor polynomial, include terms through the stated order, including mixed terms when required.
  • For a linearization at \((a,b)\), use \(h=x-a\) and \(k=y-b\), with derivatives evaluated at \((a,b)\).
  • Call \(df\) a predicted change unless exact equality has been established.

Why These Distinctions Matter

These checks separate different mathematical claims. Partial derivatives describe coordinate-direction behavior; differentiability controls the approximation in all directions. A linearization keeps first-order contributions; a second-order approximation also accounts for quadratic and mixed contributions. A differential estimates a change, while direct substitution determines the exact change.

When an answer seems unreasonable, check the base point and increments first, then check whether the approximation order matches the terms retained. Finally, distinguish the value predicted by the approximation from the function value itself. These habits make local calculations more reliable and clarify exactly what a result guarantees.

Check Your Understanding

For each question, identify the relevant condition or calculation rather than relying only on a memorized formula.

  1. Why do the existence of \(f_x(a,b)\) and \(f_y(a,b)\) not by themselves establish differentiability at \((a,b)\)?
  2. For \(f(x,y)=xy\) at the origin, what is the mixed second-order term, and why is it not part of the first-order linearization?
  3. At base point \((2,-1)\), what increments correspond to target point \((2.03,-0.95)\)?
  4. State a sufficient condition involving the first partial derivatives that guarantees differentiability at a point.
  5. How does a differential differ from the exact change in function value?

Previous

Next

Back to Tutorials