Difficulty: Intermediate to Advanced | Prerequisites: basic implicit differentiation (see companion notes), quotient rule, product rule, trigonometric derivatives
Tags: second derivative implicit, normal line, concavity, quotient rule, product rule implicit, ellipse, trig implicit differentiation, advanced calculus practice
Once you can find dy/dx by implicit differentiation, the next layer of skill involves finding second derivatives, writing equations for normal lines (not just tangent lines), and handling equations that involve products of x and y, trigonometric functions, or higher powers. These are the problems that separate a basic understanding of the technique from exam-ready fluency. If you are comfortable with the basic circle example from the companion notes, you are ready for this material.
To find second derivatives implicitly, differentiate y' with respect to x using the quotient (or product) rule, then substitute the expression for y' back in so the answer is in terms of x and y only. Normal lines are perpendicular to tangent lines, so flip the tangent slope and negate it. Curves involving xy products need the product rule during differentiation, and trig functions need their own derivative rules on top of the chain rule.
Normal line
The line perpendicular to the tangent line at a given point on a curve. If the tangent has slope m, the normal has slope -1/m. Think of it as the line that points straight into the curve rather than along it.
Concavity
Whether a curve bends upward (concave up, y'' > 0) or downward (concave down, y'' < 0) at a point. The second derivative tells you this. In simple terms, concave up looks like a cup, concave down looks like a cap.
Second implicit derivative (y'')
The derivative of dy/dx, found by differentiating dy/dx with respect to x while still treating y as a function of x. The result should contain only x and y (not y'), because you substitute the first derivative back in.
Product rule in implicit context
When an equation contains a term like xy, differentiating it with respect to x requires the product rule: d/dx[xy] = x · dy/dx + y · 1. Think of it as: both x and y are moving, so each takes a turn being differentiated while the other stays put.
The source material demonstrates this with x² + y² = 1 (the unit circle). The procedure generalises to any implicit equation.
Start with the first derivative. For x² + y² = 1: differentiate to get 2x + 2y · y' = 0, so y' = -x/y.
Differentiate y' = -x/y with respect to x using the quotient rule.
Quotient rule: d/dx[u/v] = (v · du/dx - u · dv/dx) / v².
Here u = -x, v = y, so du/dx = -1 and dv/dx = y'.
y'' = (y(-1) - (-x)(y')) / y² = (-y + x · y') / y².
Substitute y' = -x/y back in to eliminate y' from the expression.
y'' = (-y + x(-x/y)) / y² = (-y - x²/y) / y² = -(y² + x²) / y³.
Simplify using the original equation. Since x² + y² = 1: y'' = -1/y³.
The normal line at a point on a curve is perpendicular to the tangent line at that point.
Find dy/dx at the point (this gives the tangent slope m).
The normal slope is -1/m (the negative reciprocal).
Use point-slope form with the normal slope and the given point.
For example, on x² + y² = 25 at (3, 4): the tangent slope is -3/4, so the normal slope is 4/3. The normal line is y = (4/3)(x - 3) + 4 = (4/3)x. Notice this passes through the origin, which makes sense because the normal to a circle at any point passes through the centre.
When the equation contains a term like xy or x²y, you need the product rule during differentiation.
For a term xy: d/dx[xy] = x(dy/dx) + y(1) = x · y' + y.
For a term x²y: d/dx[x²y] = x²(dy/dx) + y(2x) = x² · y' + 2xy.
After differentiating, collect all y' terms on one side, factor out y', and solve.
Worked example with x³ + y³ = 6xy:
Differentiate: 3x² + 3y² · y' = 6(x · y' + y).
Expand the right side: 3x² + 3y² · y' = 6x · y' + 6y.
Collect y' terms: 3y² · y' - 6x · y' = 6y - 3x².
Factor: y'(3y² - 6x) = 6y - 3x².
Solve: y' = (6y - 3x²) / (3y² - 6x) = (2y - x²) / (y² - 2x).
When y appears inside a trig function, the chain rule applies in the same way.
d/dx[sin(y)] = cos(y) · dy/dx.
d/dx[cos(y)] = -sin(y) · dy/dx.
d/dx[tan(y)] = sec²(y) · dy/dx.
Example: sin(x) + cos(y) = 1.
Differentiate: cos(x) + (-sin(y)) · y' = 0.
Solve: y' = cos(x) / sin(y).
\frac{d}{dx}[xy] = x\frac{dy}{dx} + y\frac{d}{dx}[x^2 y] = x^2 \frac{dy}{dx} + 2xy\text{Normal slope} = -\frac{1}{m_{\text{tangent}}}\frac{d}{dx}[\sin(y)] = \cos(y) \cdot \frac{dy}{dx}y'' = \frac{d}{dx}\left[\frac{dy}{dx}\right] \text{ (substitute } y' \text{ back in at the end)}Students often apply the product rule to xy but forget to include the dy/dx factor on the y term. The correct differentiation of xy is x(dy/dx) + y, not x + y.
When finding y'', students frequently leave the answer in terms of y' instead of substituting back. The final answer for y'' should contain only x and y.
Students sometimes think the normal line and the tangent line are the same thing. They are perpendicular to each other. If one has slope m, the other has slope -1/m.
On trig problems, students occasionally write d/dx[sin(y)] = cos(y), omitting the chain rule factor dy/dx. The correct result is cos(y) · dy/dx.
⚠️ "Find the tangent and normal lines" is a common two-part exam question. Full marks require both equations, and students who skip the normal line lose easy points.
⚠️ Curves with xy terms (like the folium of Descartes, x³ + y³ = 6xy) are exam favourites because they test whether you can apply the product rule and the chain rule simultaneously.
⚠️ Second derivative problems via implicit differentiation test three skills at once: the quotient rule, the chain rule, and algebraic substitution. These appear on harder exams and are worth practising until the back-substitution step feels automatic.
⚠️ Trig-based implicit equations (sin(xy) = x, e^y + x = y, etc.) are common on final exams. They combine implicit differentiation with trig or exponential derivative rules.
Fill in the blank: d/dx[xy] = ___.
True or false: The normal line to a curve has the same slope as the tangent line. ___
Fill in the blank: d/dx[sin(y)] = ___.
True or false: When finding y'', you should leave your answer in terms of x, y, and y'. ___
Fill in the blank: If the tangent slope at a point is 2, the normal slope is ___.
Answers: 1. x(dy/dx) + y. 2. False (it has slope -1/m, the negative reciprocal). 3. cos(y) · dy/dx. 4. False (substitute y' back so the answer contains only x and y). 5. -1/2.
Q: Find dy/dx for the ellipse x²/9 + y²/4 = 1.
A: Differentiate: 2x/9 + (2y/4)(dy/dx) = 0, so (2y/4)(dy/dx) = -2x/9. Solve: dy/dx = -4x/(9y).
Q: Find the equation of the normal line to x² + y² = 25 at the point (3, 4).
A: The tangent slope is -3/4, so the normal slope is 4/3. Using point-slope form: y = (4/3)(x - 3) + 4 = (4/3)x.
Q: Find dy/dx for x³ + y³ = 6xy.
A: Differentiate using the chain rule on y³ and the product rule on 6xy: 3x² + 3y²(dy/dx) = 6x(dy/dx) + 6y. Rearrange: (3y² - 6x)(dy/dx) = 6y - 3x². Solve: dy/dx = (2y - x²)/(y² - 2x).
Q: Find dy/dx for sin(x) + cos(y) = 1.
A: Differentiate: cos(x) - sin(y)(dy/dx) = 0. Solve: dy/dx = cos(x)/sin(y).
Q: Find y'' for x² + y² = 1 (the unit circle).
A: First, y' = -x/y. Differentiate using the quotient rule: y'' = (-y + x · y')/y². Substitute y' = -x/y: y'' = (-y - x²/y)/y² = -(y² + x²)/y³. Since x² + y² = 1: y'' = -1/y³.
Q: Find dy/dx for xy + y² = 1.
A: Apply the product rule to xy: x(dy/dx) + y + 2y(dy/dx) = 0. Factor: (x + 2y)(dy/dx) = -y. Solve: dy/dx = -y/(x + 2y).
Q: At what point(s) on the ellipse x²/9 + y²/4 = 1 is the tangent line horizontal?
A: The tangent is horizontal where dy/dx = 0. Since dy/dx = -4x/(9y), the numerator is zero when x = 0. Substituting x = 0 into the ellipse equation: y²/4 = 1, so y = ±2. The tangent is horizontal at (0, 2) and (0, -2).
Q: For the curve xy = 6, find the tangent and normal line equations at (2, 3).
A: Differentiate: x(dy/dx) + y = 0, so dy/dx = -y/x = -3/2. Tangent: y = (-3/2)(x - 2) + 3 = (-3/2)x + 6. Normal slope = 2/3. Normal: y = (2/3)(x - 2) + 3 = (2/3)x + 5/3.
The product rule technique used on xy terms is exactly the same skill needed for related rates, where you differentiate equations with respect to time and every variable picks up a d/dt factor.
Second derivatives connect to concavity and the second derivative test for local extrema. Being able to find y'' implicitly lets you determine the shape of curves that cannot be expressed as y = f(x).
Normal lines appear in physics (forces perpendicular to surfaces, reflection angles) and in multivariable calculus (gradient vectors, which point in the normal direction to level curves).
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