Why Differentiate the Implicit Relation Again?
An equation relating \(x\) and \(y\) can define \(y\) as a function of \(x\) near a point, even when solving the equation explicitly for \(y\) is inconvenient. The first implicit derivative gives the slope of that local function. Differentiating once more gives its curvature: the rate at which the slope changes as \(x\) changes.
The second differentiation requires care. The partial derivatives of the defining relation are functions of both variables, and along the curve both \(x\) and \(y\) vary. The chain rule therefore contributes terms that are easy to omit. We derive the formula from the first implicit differentiation identity and make its conditions explicit.
Conditions for a Second Implicit Derivative
Let \(F\) be a function of two variables, and suppose a point \((a,b)\) lies on the relation \(F(x,y)=0\). To obtain a local function \(y=y(x)\) through that point, the relevant nonzero-partial condition is \(F_y(a,b)\ne 0\). For the second derivative formula, \(F\) must also have continuous second partial derivatives near \((a,b)\).
Every partial derivative in this formula is evaluated at \((x,y(x))\). The condition \(F_y\ne 0\) is needed to divide by \(F_y\); it also ensures the local branch is a function of \(x\) under the implicit function theorem. If this condition fails, this formula does not apply at that point.
Deriving the Formula
Along the local branch, the equation \(F(x,y(x))=0\) is an identity in \(x\). Differentiating it by the chain rule gives
This first-derivative identity is the starting point for the second derivative. Each partial derivative in it depends on both \(x\) and \(y(x)\), so differentiating \(F_x\) or \(F_y\) requires the chain rule as well.
Along the branch, its derivative is \(F_{xx}+F_{xy}y'\): the first term comes from changing \(x\), and the second from changing \(y\).
Use the product rule. The derivative of \(F_y\) along the branch is \(F_{yx}+F_{yy}y'\), and the derivative of the other factor is \(y''\).
Use equality of the mixed partials, \(F_{yx}=F_{xy}\), which holds because the second partial derivatives are continuous.
Writing out those derivatives gives
In the second line, the product rule accounts for the term \(F_y y''\). In the last line, \(F_{yx}=F_{xy}\), so the two terms involving one factor of \(y'\) combine. Since \(F_y\ne 0\), rearranging and dividing by \(F_y\) yields
This derivation also explains the factor \(2\) in the numerator: one \(F_{xy}y'\) comes from differentiating \(F_x\), and another comes from differentiating \(F_y\) along the branch.
Worked Examples
Worked Example: A Circle at a Point on Its Upper Branch
Find \(y''\) at \((3,4)\) on the circle \(x^2+y^2=25\), using implicit differentiation. Define \(F(x,y)=x^2+y^2-25\). Then \(F(3,4)=9+16-25=0\), and \(F_y=2y\), so \(F_y(3,4)=8\ne 0\). The required local branch exists and is twice differentiable there.
The first implicit derivative is \(F_x+F_y y'=0\). At the point, \(F_x=2x=6\) and \(F_y=8\), so
The second partial derivatives are \(F_{xx}=2\), \(F_{xy}=0\), and \(F_{yy}=2\). Substituting all the values into the second-derivative formula gives
Check with the explicit upper branch \(y=\sqrt{25-x^2}\). Its first derivative is \(y'=-x(25-x^2)^{-1/2}\), and differentiating gives \(y''=-(25-x^2)^{-3/2}\bigl((25-x^2)+x^2\bigr)=-25(25-x^2)^{-3/2}\). At \(x=3\), this is \(-25/4^3=-25/64\), as obtained implicitly.
Worked Example: A Relation with a Mixed Partial
Find the second derivative at \((2,2)\) on \(xy+y^2=8\). Set \(F(x,y)=xy+y^2-8\). The point is on the relation because \(2\cdot2+2^2-8=0\). Here \(F_y=x+2y\), so \(F_y(2,2)=2+4=6\ne0\).
Also, \(F_x=y\). The first derivative equation gives \(2+6y'=0\), hence \(y'=-1/3\) at the point. The needed second partial derivatives are \(F_{xx}=0\), \(F_{xy}=1\), and \(F_{yy}=2\). Therefore
For a check, solving the quadratic for the branch through \((2,2)\) gives \(y=(-x+\sqrt{x^2+32})/2\). Its first derivative is \(\bigl(-1+x/\sqrt{x^2+32}\bigr)/2\), which at \(x=2\) equals \((-1+2/6)/2=-1/3\). Differentiating again gives \(y''=16(x^2+32)^{-3/2}\). At \(x=2\), this is \(16/216=2/27\), matching the implicit result.
Worked Example: A Nonlinear Power of \(y\)
Find \(y''\) at \((1,2)\) on \(x^2+y^3=9\). Let \(F(x,y)=x^2+y^3-9\). The point satisfies \(1^2+2^3=9\), and \(F_y=3y^2\), so \(F_y(1,2)=12\ne0\).
Since \(F_x=2x\), the first derivative equation at the point is \(2+12y'=0\). Thus \(y'=-1/6\). The second partial derivatives are \(F_{xx}=2\), \(F_{xy}=0\), and \(F_{yy}=6y=12\) at \((1,2)\). Hence
To check, the branch can be written \(y=(9-x^2)^{1/3}\) near \(x=1\). Differentiating gives \(y'=-2x/(3y^2)\). Differentiating that expression along the branch gives \(y''=-2/(3y^2)+4xy'/(3y^3)\). Substituting \(x=1\), \(y=2\), and \(y'=-1/6\) gives \(-1/6-1/36=-7/36\), in agreement.
A Common Chain-Rule Error
A frequent error is to differentiate \(F_x\) as though it depended only on \(x\), writing \(dF_x/dx=F_{xx}\). Along an implicit branch, however, \(y\) changes with \(x\). The correct derivative is \(dF_x/dx=F_{xx}+F_{xy}y'\).
For example, with \(F(x,y)=xy+y^2-8\), the partial derivative \(F_x=y\). Along the branch, \(dF_x/dx=dy/dx=y'\), not \(0\). Equivalently, \(F_{xx}=0\) but \(F_{xy}y'=y'\), so the full chain-rule expression is \(0+y'\). Omitting this term in the second differentiation would lose a contribution to the numerator and give the wrong curvature.
Interpreting the Result
The value of \(y''\) describes how the slope \(y'\) changes along the branch as \(x\) changes. A positive second derivative means the slope is increasing locally; a negative second derivative means it is decreasing locally. The formula computes this change from the partial derivatives of the defining relation and the slope already found.
The formula is local, not a guarantee that the whole curve is one function of \(x\). A curve can have different branches, and where \(F_y=0\), this representation may fail even if the curve itself continues. Near such a point, the relation might instead be represented as \(x\) in terms of \(y\), or require a separate analysis. The condition \(F_y\ne0\) is therefore a substantive requirement, not a technical detail to ignore.
Summary
Starting with \(F_x+F_y y'=0\), differentiate each term along the local branch. The chain rule and product rule produce the second derivative identity; continuous second partials allow the mixed partials to be combined. Provided \(F_y\ne0\), solve for \(y''\):
Evaluate every partial derivative at \((x,y(x))\), and use the first implicit derivative to find \(y'\). Keeping the branch conditions and all chain-rule terms visible makes the calculation both reliable and interpretable.
Check Your Understanding
Use the conditions and derivation in this tutorial to answer the following questions.
- What assumptions on \(F\) and the point \((a,b)\) ensure a twice-differentiable local branch \(y=y(x)\)?
- Why does differentiating \(F_x\) along the branch produce \(F_{xx}+F_{xy}y'\), rather than just \(F_{xx}\)?
- Where do the two terms involving \(F_{xy}y'\) arise in the derivation?
- For \(x^2+y^2=25\) at \((3,4)\), what are \(y'\) and \(y''\)?
- Why can the formula not be used by dividing by \(F_y\) at a point where \(F_y=0\)?
