Second Derivatives of Implicit Curves

Why Differentiate an Implicit Curve Twice?

An equation such as \(x^2+y^2=25\) describes a plane curve without expressing \(y\) as a single formula in terms of \(x\). Near a point where the curve can be represented as a graph \(y=y(x)\), implicit differentiation gives the slope of that graph. Differentiating again gives its second derivative, which describes how the slope changes and helps determine local concavity and curvature.

The second derivative is not obtained by simply differentiating the partial derivatives of \(F\). The variable \(y\) depends on \(x\), so every derivative involving \(y\) must include the chain rule. Keeping track of this dependence is the central step.

The Second-Derivative Formula

Write \(F_x\) and \(F_y\) for the partial derivatives of \(F\), with \(F_{xx}\), \(F_{xy}\), and \(F_{yy}\) denoting its second partial derivatives. Along a local graph \(y=y(x)\), the relation \(F(x,y(x))=0\) has first derivative

If \(F_y\ne 0\), this gives the first derivative. Differentiating once more gives the second derivative:

All partial derivatives in this formula are evaluated at \((x,y(x))\). The first derivative appearing in the numerator can be replaced by \(-F_x/F_y\) wherever \(F_y\ne 0\).

Why the Formula Works

The first derivative equation contains two terms, \(F_x\) and \(F_y y'\). Each term depends on \(x\) along the curve. In particular, when differentiating a partial derivative such as \(F_x(x,y(x))\), the chain rule gives \(F_{xx}+F_{xy}y'\). Applying that rule to both terms explains every contribution in the second-derivative formula.

In full, the differentiated equation is

Combining the middle terms gives \(F_{xx}+2F_{xy}y'+F_{yy}(y')^2+F_y y''=0\). Isolating \(y''\) proves the stated formula. The factor \(2\) on the mixed partial term appears because \(F_{xy}y'\) arises once from differentiating each term of the first derivative equation.

Worked Example: A Curve with a Mixed Partial Derivative

Second Derivative and Signed Curvature

For a local graph \(y=y(x)\), the second derivative describes how the slope changes with \(x\). Its sign indicates concavity: where \(y''<0\), the graph is concave down; where \(y''>0\), it is concave up. The geometric curvature also depends on the slope, so it is not generally equal to \(y''\).

For the graph parametrized by \(\mathbf r(x)=(x,y(x))\), define its signed curvature, with this graph orientation, by \(\kappa_s=\det(\mathbf r',\mathbf r'')/\|\mathbf r'\|^3\). Here \(\mathbf r'=(1,y')\), \(\mathbf r''=(0,y'')\), and \(\|\mathbf r'\|=\sqrt{1+(y')^2}\). Therefore

The unsigned curvature is \(|\kappa_s|\). The signed value retains the concavity information for this orientation; reversing the direction used to traverse a curve can change the sign convention. To find curvature from an implicit equation, first calculate \(y'\) and \(y''\) at the point, then substitute both into the curvature formula.

When the Formula Does Not Apply

The condition \(F_y\ne0\) is essential to this formula for \(y''\): it permits division by \(F_y\) and, under the implicit function theorem's smoothness assumptions, supports a local graph \(y=y(x)\). If \(F_y=0\) at a point, this formula cannot be used there to solve for \(y''\). That does not by itself prove the curve is singular. The curve may instead be a graph \(x=x(y)\), or another representation may be needed.

Also, check that the point actually satisfies \(F(x,y)=0\) before evaluating derivatives. A derivative calculation at a point not on the curve does not describe the curve's local behavior there. Finally, evaluate the partial derivatives at the specified point only after identifying their formulas; keeping the evaluation steps visible helps catch sign and substitution errors.

A Common Mistake: Losing a Chain-Rule Term

For the circle \(x^2+y^2=25\), the first derivative equation is \(2x+2yy'=0\). A common incorrect second differentiation is \(2+2yy''=0\), which gives \(y''=-1/y\) and predicts \(-1/4\) at \((3,4)\). This result is wrong because differentiating \(2yy'\) requires the product rule:

The missing term \(2(y')^2\) matters. The corrected equation is \(2+2(y')^2+2yy''=0\). At \((3,4)\), where \(y'=-3/4\), it gives \(2+18/16+8y''=0\), hence \(8y''=-50/16\) and \(y''=-25/64\). Substitution into the original differentiated equation confirms the correction.