Differentiation Mastery II

When an Equation Defines a Function

An equation involving \(x\) and \(y\) can describe a curve without giving \(y\) explicitly in terms of \(x\). Near some points, the curve still has a unique, differentiable branch \(y=\phi(x)\). Implicit differentiation is the technique of finding the derivative of that branch directly from the equation.

The derivative of an inverse, established in the previous tutorial, is a useful guide: a nonzero derivative allows a local relation to be solved in the opposite direction. For an equation \(F(x,y)=0\), the relevant quantity is the partial derivative with respect to \(y\). But before differentiating the equation, we need to know that a local function \(\phi\) actually exists. The theorem below supplies that justification.

For the theorem, assume \(F\) is defined on an open rectangle around \((a,b)\), that its first partial derivatives exist and are continuous there, and that \(F(a,b)=0\). The condition \(F_y(a,b)\ne0\) says that, at the point in question, changing \(y\) changes \(F\) to first order. It is this condition that lets us solve locally for \(y\).

The Local Implicit Function Theorem

Proof. Since \(F_y(a,b)\ne0\) and \(F_y\) is continuous, we can choose a closed rectangle around \((a,b)\), contained in the domain of \(F\), on which \(F_y\) has the same nonzero sign as \(F_y(a,b)\). Choose \(r>0\) small enough that \(b-r\) and \(b+r\) are in this rectangle. For fixed \(x=a\), the Mean Value Theorem applied as \(y\) varies shows that \(F(a,y)\) is strictly monotone on \([b-r,b+r]\). Since \(F(a,b)=0\), the values \(F(a,b-r)\) and \(F(a,b+r)\) have opposite signs.

By continuity of \(F\), these two strict signs persist when \(x\) is sufficiently close to \(a\). Thus there is an open interval \(J\) containing \(a\) such that, for every \(x\in J\), the values \(F(x,b-r)\) and \(F(x,b+r)\) have opposite signs. The Intermediate Value Theorem gives at least one \(y\in(b-r,b+r)\) with \(F(x,y)=0\). For each fixed \(x\), the Mean Value Theorem and the nonzero constant sign of \(F_y\) show that \(y\mapsto F(x,y)\) is strictly monotone on this interval. Hence that root is unique. Call it \(\phi(x)\), and set \(K=(b-r,b+r)\). In particular, \(\phi(a)=b\).

We next show that \(\phi\) is continuous. Fix \(x_0\in J\), and write \(y_0=\phi(x_0)\). Given \(\varepsilon>0\), choose \(s\) with \(0<s<\varepsilon\) small enough that \(y_0-s,y_0+s\in K\). Strict monotonicity in \(y\) and \(F(x_0,y_0)=0\) imply that \(F(x_0,y_0-s)\) and \(F(x_0,y_0+s)\) have opposite signs. By continuity, those signs persist for \(x\) sufficiently close to \(x_0\). The unique root \(\phi(x)\) must then lie between \(y_0-s\) and \(y_0+s\). Therefore \(|\phi(x)-y_0|<s<\varepsilon\), which proves continuity at \(x_0\).

Now fix \(x\in J\) and take nonzero \(h\) sufficiently small that \(x+h\in J\). Subtract the two zero values \(F(x+h,\phi(x+h))\) and \(F(x,\phi(x))\), inserting \(F(x,\phi(x+h))\) between them. Apply the one-variable Mean Value Theorem first in the \(x\)-coordinate and then in the \(y\)-coordinate. There are points \(\xi_h\) between \(x\) and \(x+h\), and \(\eta_h\) between \(\phi(x)\) and \(\phi(x+h)\), such that

Because \(\phi\) is continuous, \(\phi(x+h)\to\phi(x)\) as \(h\to0\). Thus \(\xi_h\to x\) and \(\eta_h\to\phi(x)\). Also \(F_y(x,\phi(x))\ne0\), since \(F_y\) has a fixed nonzero sign on the chosen rectangle. Divide the displayed identity by \(hF_y(x,\eta_h)\), then let \(h\to0\). Continuity of the partial derivatives gives

This holds for every \(x\in J\), proving differentiability and the formula. Finally, strict monotonicity in \(y\) ensures that for each \(x\in J\) there is at most one root in \(K\). This proves the asserted uniqueness. \(\square\)

The theorem turns the usual formal procedure into a justified one: locally, \(y\) really is a differentiable function of \(x\). Differentiating \(F(x,\phi(x))=0\) by the chain rule then gives \(F_x+F_y\phi'=0\), which is precisely the formula proved above.

Second Derivatives of Implicit Functions

When \(F\) has continuous second partial derivatives, the first-derivative formula can be differentiated once more. The resulting expression uses only the partial derivatives of \(F\) and the already computed value of \(\phi'\).

Proof. The first-derivative formula expresses \(\phi'\) as a quotient of the functions \(F_x(x,\phi(x))\) and \(F_y(x,\phi(x))\). The second partial derivatives of \(F\) are continuous, so these compositions are continuously differentiable wherever \(\phi\) is continuously differentiable. The denominator does not vanish on the rectangle used in the first theorem. The quotient rule therefore shows that \(\phi'\) is continuously differentiable, so \(\phi\) is twice differentiable.

Differentiate \(F_x(x,\phi(x))+F_y(x,\phi(x))\phi'(x)=0\) using the chain rule and product rule. Since the mixed partial derivatives agree for a function with continuous second partial derivatives, this gives

where each partial derivative is evaluated at \((x,\phi(x))\). Combine the two mixed-partial terms, then solve for \(\phi''\). The result is the stated formula. \(\square\)

What the Nonzero Partial Derivative Does—and Does Not—Guarantee

The condition \(F_y(a,b)\ne0\) is sufficient for a unique differentiable local graph \(y=\phi(x)\). It is important not to drop it just because an equation can be manipulated or differentiated formally. At a point where \(F_y=0\), the equation might still describe a graph, but this theorem no longer guarantees one, and the displayed derivative formula cannot be used.

A common pitfall is to differentiate an equation and treat the resulting formula as proof that a branch exists. The logical order matters: first establish the local graph, then differentiate its defining identity. When \(F_y\ne0\), the theorem justifies both steps. When it vanishes, another argument is needed, and a graph may fail to exist altogether.