Difficulty: Intermediate | Prerequisites: basic derivative rules (power rule, chain rule, product rule, quotient rule)
Tags: implicit differentiation, dy/dx, chain rule, tangent line, circle equation, implicit function, calculus, derivatives
Implicit differentiation is the technique you reach for when you cannot (or do not want to) isolate y as an explicit function of x before differentiating. In single-variable calculus you are used to functions written as y = f(x), but many important curves, circles, ellipses, and higher-order relations, are defined by an equation that tangles x and y together. This topic sits at the junction of the chain rule and the algebra of equations: if you are comfortable applying the chain rule to composite functions and solving linear equations for a variable, you have everything you need. Understanding implicit differentiation also opens the door to related rates problems and to finding tangent lines on curves that explicit methods cannot handle.
When x and y are mixed together in one equation, differentiate both sides with respect to x, treat y as a function of x (so every y term picks up a dy/dx via the chain rule), then solve for dy/dx. That derivative gives you slopes, tangent lines, and rates of change on curves you could never write as y = something.
Implicit equation
An equation relating x and y where y is not isolated on one side. For example, x² + y² = 25 defines a circle but does not give y as an explicit formula. In simple terms, x and y are tangled up together and you work with the equation as-is.
Implicit differentiation
The process of differentiating both sides of an implicit equation with respect to x, applying the chain rule wherever y appears, and then solving algebraically for dy/dx. Think of it as: "differentiate everything in sight, remember y depends on x, then isolate dy/dx."
dy/dx (also written y')
The derivative of y with respect to x. In implicit differentiation this symbol starts as an unknown you are solving for, rather than something you compute directly from a formula.
Chain rule (in this context)
The rule that says: when you differentiate a function of y with respect to x, you get the derivative with respect to y, multiplied by dy/dx. For example, d/dx[y²] = 2y · (dy/dx). Think of it as: y is secretly a function of x, so differentiating anything involving y always produces a dy/dx factor.
Tangent line
A straight line that touches a curve at exactly one local point and has the same slope as the curve at that point. In the context of a circle, the tangent line at any point is perpendicular to the radius drawn to that point.
Point-slope form
The equation y - y₁ = m(x - x₁), used to write the equation of a line when you know a point (x₁, y₁) on the line and the slope m. This is the standard finishing step after implicit differentiation gives you the slope.
Start with an equation involving x and y (for example, x² + y² = 25).
Differentiate every term on both sides with respect to x.
Terms involving only x differentiate normally: d/dx[x²] = 2x.
Terms involving y require the chain rule: d/dx[y²] = 2y · dy/dx, because y itself is a function of x.
Constants differentiate to zero: d/dx[25] = 0.
Collect all terms containing dy/dx on one side of the equation.
Factor out dy/dx if it appears in more than one term.
Solve for dy/dx algebraically.
The source material walks through finding the tangent line to x² + y² = 25 at the point (3, 4).
Step 1: Differentiate both sides
Differentiating x² + y² = 25 with respect to x gives 2x + 2y · y' = 0.
Step 2: Solve for y'
Rearranging: y' = -x / y.
This formula gives the slope of the circle at any point (x, y) on it.
Step 3: Evaluate at the given point
At (3, 4): y' = -3/4.
So the tangent line has slope -3/4.
Step 4: Write the tangent line equation
Using point-slope form y = m(x - x₁) + y₁ with m = -3/4 and (x₁, y₁) = (3, 4):
y = (-3/4)(x - 3) + 4 = (-3/4)x + 25/4.
From high-school geometry, a tangent to a circle is perpendicular to the radius at the point of tangency. The radius from the origin to (3, 4) has slope 4/3. The tangent slope is -3/4. Multiplying these gives (4/3)(-3/4) = -1, which confirms perpendicularity. This is a useful sanity check on any circle-tangent problem.
Once you have y' = -x/y, you can find y'' by differentiating y' again using the quotient rule.
Applying the quotient rule to y' = -x/y:
y'' = d/dx[-x/y] = -(y · 1 - x · y') / y².
Substitute y' = -x/y back in:
y'' = -(y - x(-x/y)) / y² = -(y + x²/y) / y² = -(y² + x²) / y³.
For the circle x² + y² = 25, the numerator simplifies: y² + x² = 25, so y'' = -25/y³.
The second derivative tells you about the concavity of the curve at any point.
\frac{d}{dx}[y^n] = n\,y^{n-1} \cdot \frac{dy}{dx}\text{For } x^2 + y^2 = 25: \quad y' = \frac{-x}{y}\text{Point-slope form: } y - y_1 = m(x - x_1)\text{Second derivative: } y'' = \frac{-(y^2 + x^2)}{y^3}\text{For the circle: } y'' = \frac{-25}{y^3}The first formula above is the one to internalise. Every implicit differentiation problem is built on it: any power of y, differentiated with respect to x, drops the power by one and multiplies by dy/dx.
Students often differentiate y² and write 2y, forgetting the chain rule factor dy/dx. The correct result is 2y · (dy/dx). Every y term needs that factor.
Students sometimes try to solve for y first (e.g. y = √(25 - x²)) and then differentiate explicitly. This works for circles but fails for many implicit equations, and even where it works it introduces complications like ± signs. The whole point of implicit differentiation is to avoid that step.
When substituting back into y'' to simplify, students forget to replace y' with its expression (-x/y) and leave the second derivative in terms of both y' and y. The final answer should be in terms of x and y only.
Students sometimes confuse the slope of the radius with the slope of the tangent. On a circle centred at the origin, the radius to (a, b) has slope b/a. The tangent slope is -a/b (the negative reciprocal), not b/a.
⚠️ Implicit differentiation is almost always tested. Expect at least one problem asking for dy/dx on a curve that is not solvable for y.
⚠️ "Find the equation of the tangent line" is the most common follow-up. You need the derivative (slope) and the point, then point-slope form.
⚠️ Second derivatives via implicit differentiation appear on harder exams. The quotient rule step and the back-substitution of y' are where marks are lost.
⚠️ The perpendicularity check (radius slope times tangent slope equals -1) is a quick way to verify circle problems. Some exams ask for the normal line (which is the radius direction), so know both.
True or false: When differentiating y³ with respect to x, the result is 3y². ___
Fill in the blank: For the circle x² + y² = r², the implicit derivative is dy/dx = ___.
True or false: The tangent line to a circle is parallel to the radius at the point of tangency. ___
Fill in the blank: Point-slope form of a line is y - y₁ = ___(x - x₁).
True or false: To find y'', you differentiate y' with respect to x, treating y as a function of x. ___
Answers: 1. False (it should be 3y² · dy/dx). 2. -x/y. 3. False (perpendicular, not parallel). 4. m (the slope). 5. True.
Q: Find dy/dx for the equation x² + y² = 25.
A: Differentiate both sides: 2x + 2y(dy/dx) = 0. Solve: dy/dx = -x/y.
Q: What is the slope of the tangent line to x² + y² = 25 at the point (3, 4)?
A: Substitute into dy/dx = -x/y: slope = -3/4.
Q: Write the equation of the tangent line to x² + y² = 25 at (3, 4).
A: Using point-slope form: y = (-3/4)(x - 3) + 4, which simplifies to y = (-3/4)x + 25/4.
Q: How can you verify geometrically that your tangent line to a circle is correct?
A: Check that the tangent slope is the negative reciprocal of the radius slope. The radius from (0, 0) to (3, 4) has slope 4/3; the tangent slope -3/4 satisfies (4/3)(-3/4) = -1.
Q: Find the second derivative y'' for x² + y² = 25.
A: Start from y' = -x/y. Apply the quotient rule: y'' = -(y - x · y')/y². Substitute y' = -x/y and simplify: y'' = -(y² + x²)/y³ = -25/y³.
Q: Find dy/dx for the equation x³ + y³ = 6xy (the folium of Descartes).
A: Differentiate: 3x² + 3y²(dy/dx) = 6y + 6x(dy/dx). Collect dy/dx terms: (3y² - 6x)(dy/dx) = 6y - 3x². Solve: dy/dx = (6y - 3x²)/(3y² - 6x) = (2y - x²)/(y² - 2x).
Implicit differentiation is the foundation for related rates problems, where two or more quantities change with time and are linked by an equation. The same "differentiate both sides, apply the chain rule" logic applies, with d/dt replacing d/dx.
It also connects directly to multivariable calculus: the formula dy/dx = -(∂F/∂x)/(∂F/∂y) for a level curve F(x, y) = c is exactly what implicit differentiation produces, just written in partial-derivative notation.
The geometric insight (tangent perpendicular to radius) extends to normal lines on any curve, which matter in physics (reflection of light, forces perpendicular to surfaces) and in later courses on differential geometry.
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