Try
The equation x²+y²=25 describes a circle, not a single global function y=f(x). Yet near the point (3,4), its upper branch has a well-defined tangent. Sketch the circle, mark that point and predict the slope sign without solving for the entire branch.
Predict
If x increases slightly near (3,4), y must decrease to keep the sum of squares fixed. Predict a negative slope. Now move to (5,0). Does the tangent disappear, or does it become vertical? A formula for a finite dy/dx may fail even when a geometric tangent remains meaningful.
Experiment
Differentiate x²+y²=25 while treating y as a local function of x. The chain rule gives 2x+2y y′=0, so y′=−x/y where y is nonzero. Compute these tangent slopes and sketch short segments at each point.
| Point on circle | Finite slope from −x/y | Interpretation |
|---|
| (0,5) | 0 | horizontal top tangent |
| (3,4) | −3/4 | falling upper branch |
| (−3,4) | 3/4 | rising upper branch |
| (5,0) | undefined | investigate a vertical tangent |
Before using the rule at a proposed point, substitute it into the original equation. At (1,1), the sum of squares is 2, not 25. No derivative computation can turn that point into a point on this circle.
Understand
Implicit differentiation differentiates a relationship without first isolating the dependent variable. The derivative of y² with respect to x is 2y y′, not merely 2y, because y changes with x. Similarly the derivative of xy is y+x y′ by the product rule. On xy+y²=6, this yields y+(x+2y)y′=0, hence y′=−y/(x+2y) wherever that denominator is nonzero and the local branch is valid.
The equation itself does not always select one local branch. At a regular point where the partial rate of the equation with respect to y is nonzero, the implicit function theorem supplies a local y-as-a-function-of-x under appropriate differentiability assumptions. A full proof lies beyond this lesson, but the condition explains the denominators you encounter. If the denominator vanishes, inspect the relationship instead of automatically declaring a corner or a nonexistent curve.
At (5,0) on the circle, solve locally for x=√(25−y²). Its derivative with respect to y is −y/√(25−y²), zero at y=0. The circle’s tangent is the vertical line x=5. By contrast, y²=x² is the union of two lines y=x and y=−x. At the origin they cross with slopes 1 and −1. Differentiation gives 0=0 there, which supplies no unique tangent; it is not evidence of slope zero.
Higher derivatives also require dependencies to remain visible. Starting from x+y y′=0 on the circle, differentiate again: 1+(y′)²+y y″=0. Substitute y′=−x/y to obtain y″=−(y²+x²)/y³=−25/y³. The upper branch has negative second derivative and the lower branch positive, wherever y≠0. This calculation uses both the product rule and the original constraint.
Worked synthesis: a tilted ellipse
Now use the relationship x²+xy+y²=3 to connect all these calculations. This is a separate worked curve, not the circle above. Before differentiating, locate its horizontal and vertical extent by completing squares. The identities (x+y/2)²+3y²/4=3 and (y+x/2)²+3x²/4=3 imply −2≤y≤2 and −2≤x≤2. Sketch these bounds and mark the four candidate boundary points from the table below. Unlike the circle, the top and bottom need not lie on the y-axis.
Differentiation gives 2x+y+(x+2y)y′=0. Thus y′=−(2x+y)/(x+2y) when x+2y is nonzero. The two partial derivatives of the defining expression are 2x+y and x+2y. They vanish together only at (0,0), which is not on this ellipse. Every point of the curve is therefore regular: at least one coordinate can serve as a smooth local parameter. We use the local implicit-function theorem stated above; we are not inferring regularity merely from an attractive sketch.
For a horizontal tangent set 2x+y=0 and retain x+2y≠0. Substituting y=−2x into the original equation gives 3x²=3. For a vertical tangent instead set x+2y=0 and verify 2x+y≠0; substituting x=−2y gives 3y²=3. These conditions locate actual points, not just zeros of a numerator detached from the constraint.
| Point | 2x+y | x+2y | Tangent and geometric role |
|---|
| (−1,2) | 0 | 3 | Horizontal, highest y |
| (1,−2) | 0 | −3 | Horizontal, lowest y |
| (2,−1) | 3 | 0 | Vertical, greatest x |
| (−2,1) | −3 | 0 | Vertical, least x |
At a vertical point, use x as a function of y: dx/dy=−(x+2y)/(2x+y)=0 there. This supports a vertical tangent without asserting a finite dy/dx. The local graph y=f(x) fails across the side point even though the ellipse itself remains smooth.
Differentiate the first-derivative equation again. The result is 2+2y′+2(y′)²+(x+2y)y″=0. Put A=x+2y and B=2x+y temporarily, so y′=−B/A. The numerator after substitution contains A²−AB+B²=3(x²+xy+y²)=9. Consequently y″=−18/(x+2y)³ wherever A≠0.
At (−1,2), y″=−2/3, so the horizontal point is a strict local maximum of the local y-branch. At (1,−2), y″=2/3, giving a strict local minimum. Completing the square already bounded y between −2 and 2, so these are also the global highest and lowest values on the entire curve. A second-derivative test alone would have established only the local claim.
Solving for the two y-branches gives y=[−x±√(12−3x²)]/2. On −2<x<2, x+2y is positive on the upper branch and negative on the lower branch. The upper branch is therefore concave down throughout its open interval, and the lower branch concave up. Neither has an inflection point. The joining side points are not inflections of a local y=f(x): no such single local graph passes through the full curve there. Do not confuse switching between two branches with a concavity change along one function graph.
As a final learner action, make two sketches: the whole ellipse with tangent directions, and the upper branch alone with increasing/decreasing arrows. On the upper branch, y′ is positive before x=−1 and negative after it. Check that this agrees with its top at (−1,2), while the lower branch has its minimum at x=1. The sign of the slope and the sign of its rate of change have different geometric meanings.
Apply
For each practice problem, verify the point, differentiate every term, collect y′ terms and state restrictions before dividing. For the singular example, draw both branches and explain what the algebra fails to determine. A correct “undefined formula” response is incomplete if the question asks for geometric tangent behavior.
Unfamiliar synthesis practice
Investigate x²+2xy+2y²=4. Produce a labeled sketch or coordinate description, derive both implicit derivatives, prove regularity on the curve, and locate every horizontal and vertical tangent. Classify local y-extrema and branch concavity. State separately which extrema are global and why. Do not reuse the worked ellipse’s points or its second-derivative coefficient.
Check the synthesis argument after attempting it
Differentiating gives 2(x+y)+2(x+2y)y′=0, so y′=−(x+y)/(x+2y). Differentiating the divided first equation yields 1+2y′+2(y′)²+(x+2y)y″=0. Substitution and the original constraint simplify this to y″=−4/(x+2y)³. Both derivatives require x+2y≠0 when used as y-graph formulas.
The partial derivatives are 2(x+y) and 2(x+2y). Their only common zero is the origin, which is off the curve, so every point is regular. Horizontal tangents require x=−y; substitution gives y=±2. At (−2,2), y″=−1/2 and the branch has a strict local maximum. At (2,−2), y″=1/2 and it has a strict local minimum. Since (x+y)²+y²=4 implies −2≤y≤2, these are global y-extrema too.
Vertical tangents require x=−2y, giving y=±√2 and points (−2√2,√2), (2√2,−√2). At each, x+y=−y is nonzero. The local derivative dx/dy is zero, so the vertical tangent conclusion is justified. Solving for y gives [−x±√(8−x²)]/2 on −2√2≤x≤2√2. On the open interval, the upper branch has x+2y positive and is concave down; the lower has it negative and is concave up. There are no branch inflections. Endpoint tangent failure of the y-graph formula is not a singularity of the ellipse.
Review the response for five observable features: both differentiated equations with the chain rule; the nonzero-partial regularity argument; all four tangent points checked in the original constraint; local classifications with valid second derivatives; and a global/branch argument that distinguishes coordinate bounds from local tests. A list of correct points without these reasons is incomplete synthesis evidence. This added activity is public practice within the existing lesson, not a new completion requirement or separately graded task.
Continue
Inverse-function derivatives are another application of differentiating a relationship. The next lesson will require selecting a branch and checking a nonzero slope before taking a reciprocal rate.
