Repeated Implicit Differentiation For Third Derivatives

Why a Third Implicit Derivative Needs Care

When a curve is given by an equation involving \(x\) and \(y\), it may be possible to regard \(y\) locally as a function of \(x\) without solving the equation explicitly for \(y\). Implicit differentiation then gives the first derivative and, by differentiating again, the second derivative. To find the third derivative, differentiate that second-derivative relation once more.

The key difficulty is that \(y\), \(y'\), and \(y''\) all depend on \(x\). In particular, a term such as \(y y''\) requires the product rule, and a term such as \(y^2\) requires the chain rule. Missing one of these contributions changes the result.

The Conditions and the Second-Derivative Relation

Write the implicit equation as \(F(x,y)=0\), where \(F\) has continuous partial derivatives through order three near the point under consideration. If \(F_y\ne 0\) at that point, the implicit function theorem guarantees a local function \(y=y(x)\) satisfying the equation. This condition also allows the formulas below to be solved for \(y''\) and \(y'''\).

Here, subscripts denote partial derivatives: for example, \(F_{xy}\) means differentiate \(F\) first with respect to \(x\) and then with respect to \(y\). Every partial derivative in the formulas is evaluated at \((x,y(x))\). The first implicit derivative relation is \(F_x+F_y y'=0\). Differentiating it once gives the second-derivative relation.

The terms containing \(y'\) and \(y''\) are not constants when this equation is differentiated. Applying the chain and product rules to each term gives the third-derivative formula.

Why the Third-Derivative Formula Works

Differentiate the four terms in the second-derivative relation separately. The chain rule gives \(D(F_{xx})=F_{xxx}+F_{xxy}y'\). For the next term, use the product rule on \(F_{xy}y'\); its derivative is \((F_{xxy}+F_{xyy}y')y'+F_{xy}y''\). Including the coefficient 2 produces the contributions from that term.

For \(F_{yy}(y')^2\), both factors vary: \(D(F_{yy})=F_{xyy}+F_{yyy}y'\) and \(D((y')^2)=2y'y''\). Finally, differentiating \(F_y y''\) requires the product rule, with \(D(F_y)=F_{xy}+F_{yy}y'\).

Adding these four differentiated expressions gives the theorem's formula. The coefficients 3 arise by collecting terms that appear more than once: for example, \(F_{xy}y''\) appears twice from \(D(2F_{xy}y')\) and once from \(D(F_y y'')\). The term \(F_{yy}y'y''\) appears twice from differentiating \(F_{yy}(y')^2\) and once from differentiating \(F_y y''\). This accounts for both coefficients 3 and explains why omitting a product-rule contribution is a common source of error.

Worked Examples

A Common Product-Rule Mistake

A frequent error is to differentiate \(y y''\) as though only its second factor varied. The incorrect calculation \(D(y y'')=y y'''\) drops the derivative of \(y\). Since \(y=y(x)\), the product rule instead gives

For instance, in the circle example the incorrect version would produce \(y y'''=0\), incorrectly predicting \(y'''=0\). The missing term is \(y'y''\); after including it, the correct equation is \(3y'y''+y y'''=0\). Always treat every occurrence of \(y\), \(y'\), or \(y''\) as a function of \(x\) when differentiating.

A Reliable Differentiation Routine

For a third implicit derivative, it helps to keep the calculation in stages rather than expanding everything at once.

Why This Method Is Useful

Repeated implicit differentiation gives derivative information without requiring an explicit formula for \(y\). This is useful when an equation defines a local branch naturally but solving for \(y\) would be awkward or would obscure which branch is being considered. The second and third derivatives describe how the slope changes along that branch; the same process extends to higher derivatives when the relation is sufficiently smooth.

The main discipline is to preserve the dependence \(y=y(x)\) at every step. Before substituting numerical values, write the full differentiated equation. Doing so keeps chain-rule terms visible, makes coefficients easier to check, and helps distinguish a genuine derivative from an expression that has accidentally treated \(y\) as constant.