Concavity Along an Implicit Curve
An equation such as \(x^2+y^2=25\) describes a curve without necessarily giving \(y\) as a single formula in terms of \(x\). Near a suitable point, however, part of the curve may be represented as a function \(y=y(x)\). On that branch, the second derivative \(y''\) tells us whether the graph bends upward or downward as \(x\) increases.
The key is to differentiate the defining equation while treating \(y\) as a function of \(x\). The resulting expression for \(y''\) may involve \(x\), \(y\), and \(y'\), so their signs can all matter. The equation must also define a differentiable local branch at the point under consideration; implicit differentiation does not remove that condition.
The Second-Derivative Formula
Suppose a curve is given by \(F(x,y)=0\), where \(F\) has continuous second partial derivatives near a point \((x,y)\). If \(F_y(x,y)\ne 0\), the implicit function theorem guarantees that the curve can locally be written as a twice-differentiable function \(y=y(x)\). Here \(F_x\) and \(F_y\) denote partial derivatives with respect to the indicated variables.
Differentiate the relation \(F(x,y(x))=0\) once using the chain rule. Differentiate a second time, remembering that \(y\) depends on \(x\).
When the second partial derivatives are continuous, \(F_{xy}=F_{yx}\). Thus the second equation becomes the following formula.
Why the Formula Works
The first differentiation accounts for how the point \((x,y(x))\) moves along the curve: changing \(x\) also changes \(y\). For instance, the derivative of \(F_y(x,y(x))\) is \(F_{yx}+F_{yy}y'\), not just \(F_{yy}\). Applying the product rule to \(F_y y'\) then contributes both a term involving \((y')^2\) and a term involving \(y''\).
Rearranging the second differentiated equation isolates \(y''\). Dividing by \(F_y\) is valid because the branch condition requires \(F_y\ne0\). This also explains why the denominator's sign matters: a positive numerator does not by itself determine whether \(y''\) is positive or negative.
Worked Examples
Worked Example: Concavity on the Upper Half of a Circle
Find the first and second derivatives of the upper branch of \(x^2+y^2=25\) at \((3,4)\), and classify its concavity there.
Set \(F(x,y)=x^2+y^2-25\). Then \(F_y=2y\), which equals \(8\) at \((3,4)\), so the curve is locally a function \(y(x)\) there. Differentiate the equation once:
At the given point, \(y'=-3/4\). Differentiate \(x+y y'=0\) once more:
Substitute the point and slope, showing the intermediate values:
Therefore the upper branch is concave down at \((3,4)\). The result is also checked by differentiating the first-derivative expression: \(y'=-x/y\) gives \(y''=-1/y+xy'/y^2\), which at \((3,4)\) is \(-1/4+3(-3/4)/16=-1/4-9/64=-25/64\).
Worked Example: A Positive Product Gives Concavity That Depends on the Sign of \(x\)
For the curve \(xy=12\), determine the concavity at \((2,6)\) and at \((-2,-6)\). In both cases, check the coordinates lie on the curve.
At \((2,6)\), \(xy=2\cdot6=12\). Differentiate \(xy=12\), treating \(y\) as a function of \(x\):
At \((2,6)\), \(y'=-6/2=-3\). Differentiate \(y+x y'=0\) again:
Thus \(y''=-2(-3)/2=3>0\), so the branch is concave up there. The sign pattern is \(x>0\), \(y>0\), and \(y'<0\); in this formula, negative slope divided by positive \(x\), with the leading minus sign, produces positive curvature.
At \((-2,-6)\), the check is \((-2)(-6)=12\). The slope is \(y'=-(-6)/(-2)=-3\), and the second derivative is
This branch is concave down at \((-2,-6)\). Here \(x<0\), \(y<0\), and \(y'<0\). The same negative slope now divided by negative \(x\), followed by the leading minus sign, gives a negative second derivative. As a check, solving the relation locally gives \(y=12/x\); differentiating yields \(y'=-12/x^2\) and \(y''=24/x^3\), which gives \(3\) at \(x=2\) and \(-3\) at \(x=-2\).
Worked Example: The Sign of \(y\) Reverses the Concavity
Consider \(y^2=x^3\) at the points \((4,8)\) and \((4,-8)\). Determine each branch's first and second derivatives and its concavity.
Both points satisfy the equation: \(8^2=64=4^3\) and \((-8)^2=64=4^3\). Differentiate once:
At \((4,8)\), \(y'=3(16)/(2\cdot8)=48/16=3\). At \((4,-8)\), \(y'=48/(-16)=-3\). Differentiate \(2y y'=3x^2\) again:
At \((4,8)\), the numerator is \(3(4)-3^2=12-9=3\), so \(y''=3/8>0\): the upper branch is concave up. At \((4,-8)\), the numerator is also \(12-(-3)^2=12-9=3\), but the denominator is negative, giving \(y''=3/(-8)=-3/8<0\): the lower branch is concave down.
For a direct check, the branches near \(x=4\) are \(y=x^{3/2}\) and \(y=-x^{3/2}\), since \(x\) is positive there. Their derivatives are respectively \(3\sqrt{x}/2\), \(3/(4\sqrt{x})\) and \(-3\sqrt{x}/2\), \(-3/(4\sqrt{x})\). At \(x=4\), these give slopes \(3,-3\) and second derivatives \(3/8,-3/8\), as computed implicitly.
Reading the Signs Carefully
Implicit equations do not all produce the same sign rule. In the circle example, the formula \(y''=-(1+(y')^2)/y\) has a numerator that is always positive, so the sign of \(y\) decides the sign of \(y''\). In the product example, \(y''=-2y'/x\), so the signs of both \(y'\) and \(x\) matter. In the final example, the numerator and denominator must both be evaluated.
Consequently, do not infer concavity just from a sketch, the sign of one coordinate, or the direction of the tangent. First derive the correct expression for the particular curve, then evaluate every term at the point or throughout the interval of interest.
A Reliable Procedure
For \(F(x,y)=0\), verify that \(F_y\ne0\) at the point if you want a local function \(y=y(x)\) justified by the implicit function theorem.
Use the chain rule and solve \(F_x+F_y y'=0\) for the slope when possible.
Apply both the chain rule and product rule to terms containing \(y\) or \(y'\), then isolate \(y''\).
Insert the point and slope, track the signs of all terms, and use \(y''>0\) for concave up or \(y''<0\) for concave down.
Differentiate the first-derivative relation again, or compare with an explicit local expression if one is available.
Common Pitfall: Treating \(y\) as Constant
When differentiating \(y^2=x^3\), a common error is to write \(2y=3x^2\). That calculation silently treats \(y\) as constant with respect to \(x\). Along a curve branch, \(y\) varies with \(x\), so the chain rule gives \(d(y^2)/dx=2y y'\), not \(2y\).
The correct first derivative is \(2y y'=3x^2\). Omitting \(y'\) changes the slope and makes every subsequent curvature calculation unreliable. Whenever an expression contains \(y\), explicitly account for its dependence on \(x\) while differentiating.
Why Implicit Concavity Matters
Many curves are easier to describe by a relation than by a single formula \(y=f(x)\). Implicit differentiation lets us study the slope and bending of an appropriate local branch without first solving the equation explicitly. It also makes branch differences visible: the same equation can contain points with different signs of \(y\), and those signs can produce opposite concavity.
The essential habit is to keep the whole calculation connected: establish that a local branch is available, find \(y'\), differentiate once more with the chain and product rules, and evaluate the resulting expression with its signs intact.
Check Your Understanding
For each question, show the differentiation steps and state any local branch condition that is needed.
- For \(x^2+y^2=25\), find \(y'\) and \(y''\) at \((4,3)\). Is the branch concave up or down there?
- For \(xy=8\), find \(y''\) at \((2,4)\) and at \((-2,-4)\). How do the signs of \(x\) and \(y'\) affect the results?
- For \(y^2=x^3\), find \(y'\) and \(y''\) at \((1,-1)\). Explain why the sign of \(y\) matters in the second-derivative formula.
- Why is \(y''=0\) at a point not enough, by itself, to prove that the point is an inflection point?
- State the condition involving \(F_y\) that justifies representing \(F(x,y)=0\) locally as \(y=y(x)\).
