Differentiation chain rule and implicit curves

Temperature, pressure and geometric constraints depend on several coordinates. Partial derivatives isolate one coordinate effect, but a moving observer experiences their combined effect. Total derivatives and the chain rule combine those changes correctly. Implicit functions explain when a constraint can locally be solved for one coordinate.

1 Limits require all approaches

For $f:U\subset\mathbb R^n\to\mathbb R$, assume $a$ is an accumulation point of $U\setminus\{a\}$. A limit at $a$ requires $f(x)$ to approach one number whenever $0<\|x-a\|\to0$ within the domain. Different path limits disprove existence; agreement along a few paths does not prove it.

For $f(x,y)=xy/(x^2+y^2)$ away from zero, the axes give limit zero but $y=x$ gives $1/2$, so no limit exists. By contrast,

$$\left|\frac{x^2y}{x^2+y^2}\right|\le|y|\le\sqrt{x^2+y^2},$$

which proves limit zero at the origin by a bound valid for every approach. Defining the function to be zero there makes it continuous, not automatically differentiable.

2 Total derivatives and tangent planes

For the full $\mathbb R^n$ derivative here, assume $a$ is an interior domain point and $f$ is defined on an open neighborhood of $a$. Differentiability at $a$ means $f(a+h)=f(a)+Lh+o(\|h\|)$ for a linear map $L$. For scalar output $Lh=\nabla f(a)\cdot h$. Continuous partial derivatives in a neighborhood suffice: split the increment into coordinate steps, apply the one-variable mean value theorem and control the intermediate derivative values by continuity.

The tangent plane to $z=f(x,y)$ at $(a,b,f(a,b))$ is

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

A unit direction $u$ has directional derivative $\nabla f\cdot u$. For a moving point $r(t)$, the chain rule gives $(f\circ r)'=\nabla f(r(t))\cdot r'(t)$. For vector maps it is Jacobian multiplication. This follows by substituting one local linear approximation into the other; the inner increment is $O(\|h\|)$ so the composed remainder stays $o(\|h\|)$.

Example. Let $T(x,y,z)=x^2+y^2+2z$ and $r(t)=(\cos t,\sin t,t)$. Then $T(r(t))=1+2t$, giving rate 2. The gradient method gives $(2\cos t,2\sin t,2)\cdot(-\sin t,\cos t,1)=2$. Coordinate motion can cancel two apparent temperature effects.

3 A precise implicit-function result

Theorem for a plane curve. Suppose $F$ is $C^1$ near $(a,b)$, $F(a,b)=0$, and $F_y(a,b)\ne0$. Then in suitable neighborhoods, the solutions of $F(x,y)=0$ are exactly $y=g(x)$ for a unique $C^1$ function with $g(a)=b$, and

$$g'(x)=-\frac{F_x(x,g(x))}{F_y(x,g(x))}.$$

Proof. After changing sign if necessary, continuity gives $F_y\ge m>0$ on a small rectangle. On its central vertical line, the values just below and above $b$ have opposite signs by the mean value theorem. Continuity preserves those signs for $x$ close to $a$. The intermediate value theorem gives a root between them for each such $x$, and strict increase in $y$ makes it unique. For nearby $x,x'$, the mean value theorem in $y$ gives

$$m|g(x')-g(x)|\le|F(x',g(x))-F(x,g(x))|.$$

The right side tends to zero, proving continuity, and bounded $F_x$ makes $g(x')-g(x)=O(|x'-x|)$. The differentiability expansion of $F$ at $(x,g(x))$ in the displacement $(h,g(x+h)-g(x))$ now gives

$$0=F_xh+F_y[g(x+h)-g(x)]+o(|h|).$$

Divide by $h$ and solve using $F_y\ne0$. Continuity of the derivative formula proves $g\in C^1$. $\square$

For $x^2+y^2=1$, near $(0,1)$ the theorem gives $y=\sqrt{1-x^2}$ and $g'=-x/y$. Near $(1,0)$ its $F_y$ condition fails, and the circle is not a single differentiable $y$-graph there. It can instead be written as an $x$-graph because $F_x\ne0$. Failure of one sufficient condition is not a universal claim that no implicit representation exists.

The error estimates behind the derivative rules

For the continuous-partial-derivative criterion, choose a small open box about $a$ whose closure lies in the domain. Change the coordinates one at a time from $a$ to $a+h$. The scalar mean value theorem writes the $j$th change as $f_{x_j}(a+\theta_j)h_j$, where $\|\theta_j\|\le\|h\|$. Subtracting $\sum_jf_{x_j}(a)h_j$ bounds the error by $\varepsilon\sum_j|h_j|\le\varepsilon\sqrt n\|h\|$ once all the partial derivatives differ from their values at $a$ by at most $\varepsilon$. This proves the required little-o estimate uniformly over directions. Apply it componentwise for vector outputs.

For the chain rule write $G(a+h)=G(a)+Bh+r(h)$ with $r(h)=o(\|h\|)$, and $F(b+k)=F(b)+Ak+s(k)$ with $s(k)=o(\|k\|)$, where $b=G(a)$. Then $k=Bh+r(h)=O(\|h\|)$, so the composite error $Ar(h)+s(k)$ is $o(\|h\|)$. If $k=0$, its second term is exactly zero; otherwise the product $\|s(k)\|/\|k\|$ times $\|k\|/\|h\|$ tends to zero. Thus the derivative is $AB$, in that order.

In the plane implicit-function proof, the small rectangle may be chosen closed inside the original $C^1$ neighborhood. Its compactness bounds $F_x$, while its positive lower bound for $F_y$ holds on every vertical segment used. Shrinking the horizontal interval preserves the two opposite boundary signs. These choices justify the uniform estimates used in the proof without requiring a second derivative.

4 Graded practice

E1 Foundation. Find the tangent plane to $z=x^2y+e^y$ at $(1,0,1)$.

E2 Chain rule. With $u=x+y$, $v=x-y$ and $f(u,v)=u^2+3v^2$, calculate $\partial f/\partial x$ and $\partial f/\partial y$.

E3 Implicit derivative. Near $(1,1)$, justify solving $x^2+xy+y^2=3$ for $y$ and find $dy/dx$ there.

E4 Limitation. For $F(x,y)=y^2-x^2$, explain why the theorem does not give a unique $y$-graph near zero and describe the actual solution set.

5 Full solutions

S1. $f_x=2xy$, $f_y=x^2+e^y$, so at $(1,0)$ they are 0 and 2. The plane is $z=1+2y$.

S2. $f_x=2u+6v=8x-4y$ and $f_y=2u-6v=-4x+8y$. Direct expansion $4x^2-4xy+4y^2$ verifies both.

S3. $F_y=x+2y=3\ne0$ there, so the theorem applies. $F_x=2x+y=3$, giving slope $-1$.

S4. $F_y(0,0)=0$. The solution is the union of the two lines $y=x$ and $y=-x$, giving two different $y$ values for each sufficiently small nonzero $x$. Thus no single graph describes all nearby solutions.

Summary

Multivariable limits need all approaches; differentiability needs one consistent linear approximation. Jacobian multiplication combines coordinate changes. A nonzero partial derivative in a constraint gives a local implicit graph, with a derivative formula whose validity follows from existence and uniqueness rather than formal division alone.