Second Implicit Differentiation

Why Differentiate an Implicit Equation Twice?

An equation such as \(x^2+y^2=25\) relates \(x\) and \(y\) without writing \(y\) explicitly as a function of \(x\). Near a point on the circle, however, one part of the curve can be represented by a differentiable function \(y=y(x)\). Implicit differentiation finds its derivatives directly from the equation.

To find the second derivative, differentiate the first-derivative equation again with respect to \(x\). The important detail is that \(y\) depends on \(x\), so every term involving \(y\), \(y'\), or a function of \(y\) must be differentiated using the chain rule and product rule as appropriate.

The General Setup

Suppose a curve is described by \(F(x,y)=0\), and near the point of interest it has a twice-differentiable branch \(y=y(x)\). Here \(F_x\), \(F_y\), and the other subscripts denote partial derivatives of \(F\), calculated by treating the other variable as constant. For example, \(F_x\) means differentiate with respect to \(x\) while holding \(y\) fixed.

After substituting \(y=y(x)\), the chain rule gives the first derivative of the equation:

If \(F_y\ne 0\) at the point, this equation can be solved there for \(y'\). To find \(y''\), differentiate the entire equation again with respect to \(x\). In doing so, \(F_x\) and \(F_y\) are themselves functions of both \(x\) and \(y(x)\), so each requires the chain rule.

A General Formula for the Second Derivative

Differentiate \(F_x+F_y y'=0\) term by term. The derivative of \(F_x(x,y(x))\) is \(F_{xx}+F_{xy}y'\). For the product \(F_y y'\), use the product rule; the derivative of \(F_y(x,y(x))\) is \(F_{yx}+F_{yy}y'\).

For a function \(F\) with continuous second partial derivatives, the mixed partials agree: \(F_{xy}=F_{yx}\). Combining those terms gives the standard formula.

The formula records the same terms produced by differentiating twice directly: a second partial derivative with respect to \(x\), two mixed-partial contributions, and a second partial derivative with respect to \(y\) multiplied by \((y')^2\). In practice, differentiating the first-derivative equation term by term is often the clearest way to avoid dropping a term.

A Reliable Procedure

Worked Examples

A Common Mistake: Treating \(y'\) as Constant

For the circle \(x^2+y^2=25\), the first derivative can be written \(x+y y'=0\). A common error is to differentiate \(y y'\) as though \(y\) were the only changing factor, or to treat \(y'\) as constant. For example, writing \(1+y''=0\) incorrectly gives \(y''=-1\).

The correction is to apply the product rule to \(y y'\). The derivative of the first factor \(y\) is \(y'\), and the derivative of the second factor \(y'\) is \(y''\):

Thus the correct differentiated equation is \(1+(y')^2+y y''=0\), not \(1+y''=0\). At \((3,4)\), it yields \(y''=-25/64\), as checked above. The missing \((y')^2\) and \(y\) factors are consequences of the chain and product rules, not optional details.

Why the Method Is Useful

Implicit equations often describe curves more simply than an explicit formula for \(y\). Second implicit differentiation provides \(y''\) without first solving the equation for \(y\). That can be especially useful when an equation has several branches or when solving explicitly would require complicated algebra.

The method also gives a systematic check on the work: differentiate the original relation once to find \(y'\), differentiate that equation again with every dependent factor intact, and substitute into the resulting equation. The second derivative is not obtained by differentiating only the visible powers of \(x\); it depends on how \(y\) changes along the curve.