Implicit Differentiation

§ 01What Is Implicit Differentiation?

Explicit differentiation, the kind you have done so far, starts with a function y defined entirely in terms of x, like y = x³ + sin x. You differentiate the right-hand side and you are done. But many important curves are defined by an equation relating x and y that cannot be rearranged to give y alone.

Consider the unit circle: x² + y² = 1. There is no single explicit formula y = f(x) for this curve, because for each x-value between −1 and 1, there are two y-values (one on the top half, one on the bottom). The curve is defined implicitly, x and y are related by an equation, but neither is a simple function of the other.

Implicit differentiation is the technique that lets you find dy/dx directly from such an equation, without ever solving for y. The core idea is simple but requires careful use of the chain rule: every time you differentiate a term containing y, you treat y as a function of x and apply the chain rule, which means you multiply by dy/dx.

The Core Rule of Implicit Differentiation

Whenever you differentiate yⁿ with respect to x:

d/dx[yⁿ] = n · yⁿ⁻¹ · (dy/dx)

The extra factor dy/dx comes from the chain rule — y is the inner function, x is the independent variable. This factor must appear for every y-term differentiated.

Once you have differentiated both sides, dy/dx appears in the equation. You then collect all dy/dx terms on one side and all other terms on the other, and solve algebraically for dy/dx. The result will typically be expressed in terms of both x and y: that is perfectly correct and expected.

✦ Why Does the Result Contain Both x and y? Unlike explicit differentiation where dy/dx is a function of x alone, implicit differentiation gives dy/dx in terms of both x and y. This is not an error — it is correct. To evaluate the gradient at a specific point, you substitute both the x-coordinate and the y-coordinate of that point. The curve is defined by an equation in x and y, so the gradient naturally depends on both.

§ 02Explicit vs Implicit — A Direct Comparison

The best way to understand implicit differentiation is to see the same equation treated both ways, so the contrast is clear.

Consider the equation x² + y² = 25 (a circle of radius 5). We can differentiate this implicitly in one line, or we can first solve for y and differentiate explicitly. Both should give the same dy/dx.

Explicit Approach (top half of circle)

Solve: y = √(25 − x²) = (25 − x²)^½

dy/dx = ½(25−x²)^(−½) · (−2x)

= −x ÷ √(25−x²)

= −x ÷ y (since y = √(25−x²))

Implicit Approach (whole circle)

Differentiate both sides w.r.t. x:

d/dx[x²] + d/dx[y²] = d/dx[25]

2x + 2y·(dy/dx) = 0

dy/dx = −x ÷ y ✓

Both methods give exactly the same answer: dy/dx = −x/y. But the implicit approach is two lines versus five, and it handles the entire circle (both the top and bottom halves) in a single calculation. For more complicated equations, the explicit approach either becomes extremely messy or is entirely impossible, which is why implicit differentiation is essential.

§ 03Why the Method Works — The Chain Rule Argument

Implicit differentiation is not a separate rule: it is a direct application of the chain rule. Understanding why it works is more valuable than memorising a procedure.

Suppose x and y are related by some equation F(x, y) = 0, and suppose that near a point on the curve, y really is a differentiable function of x, it is just that we cannot write down an explicit formula for it. Call this hidden function y = h(x).

Why d/dx[y²] = 2y · dy/dx

We want to compute d/dx[y²]. Since y = h(x) for some function h, we are really computing:

d/dx[(h(x))²]

This is a composite function: the outer function is (·)², and the inner function is h(x). Apply the chain rule:

d/dx[(h(x))²] = 2 · h(x) · h′(x)

Substitute back y = h(x) and dy/dx = h′(x):

= 2y · (dy/dx) ✓ The factor dy/dx appears because y is itself a function of x. Every time you differentiate any expression in y with respect to x, you must multiply by dy/dx by the chain rule.

This generalises immediately to any power or any function of y:

d/dx[yⁿ] = n·yⁿ⁻¹ · dy/dx d/dx[sin y] = cos y · dy/dx d/dx[eʸ] = eʸ · dy/dx d/dx[ln y] = (1/y) · dy/dx In every case the pattern is the same: differentiate the function of y as if y were the variable, then multiply by dy/dx.

Differentiating Terms Involving Both x and y

When a term contains both x and y multiplied together, such as xy or x²y³, you must use the product rule as well as the chain rule on the y factor.

Differentiating xy and x²y with respect to x

d/dx[xy] = d/dx[x] · y + x · d/dx[y]
= 1 · y + x · dy/dx
= y + x(dy/dx)

d/dx[x²y] = d/dx[x²] · y + x² · d/dx[y]
= 2xy + x²(dy/dx)

§ 04The Four-Step Method

A consistent routine that works for every implicit differentiation problem, however complex.

Step 1: Differentiate both sides of the equation with respect to x. For every term involving y, apply the chain rule: multiply by dy/dx.
Step 2: Expand and simplify where possible. Be careful with product rule terms.
Step 3: Collect all terms containing dy/dx on the left-hand side; move all other terms to the right-hand side.
Step 4: Factorise dy/dx from the left-hand side and divide to isolate dy/dx.

Worked example — 3x² + 2xy + y³ = 10

Step 1: Differentiate both sides w.r.t. x:
6x + 2y + 2x(dy/dx) + 3y²(dy/dx) = 0

Step 2: Already expanded.

Step 3: Collect dy/dx terms on the left:
2x(dy/dx) + 3y²(dy/dx) = −6x − 2y

Step 4: Factorise and divide:
dy/dx · (2x + 3y²) = −6x − 2y
dy/dx = (−6x − 2y) ÷ (2x + 3y²)

§ 05Common Mistakes

These are the errors that appear most often in implicit differentiation. Each has a specific cause and a specific fix.

⚠ Mistake 1 — Forgetting the dy/dx Factor When Differentiating y-Terms

When you differentiate y² with respect to x and write 2y without the dy/dx factor, you are treating y as a constant, not as a function of x. This is the most common error and it produces an expression that is not even dimensionally correct.

Wrong: d/dx[y²] = 2y

Correct: d/dx[y²] = 2y · (dy/dx) ✓

⚠ Mistake 2 — Forgetting the Product Rule on Mixed xy Terms

A term like 3xy contains both x and y. When you differentiate it with respect to x, both factors change (x changes by itself, y changes because it is a function of x). The product rule is mandatory here.

Wrong: d/dx[3xy] = 3y

Correct: d/dx[3xy] = 3y + 3x(dy/dx) ✓

⚠ Mistake 3 — Incorrect Algebra When Collecting dy/dx Terms

After differentiating, students sometimes fail to factorise dy/dx correctly. If dy/dx appears in two terms, say, 4(dy/dx) + y²(dy/dx) = 2x, you must factor it as dy/dx(4 + y²) = 2x before dividing. You cannot treat each dy/dx term independently.

Wrong: dy/dx = 2x/4 + 2x/y² (dividing each term separately: this is algebraically invalid)

Correct: dy/dx = 2x ÷ (4 + y²) (factorising first, then dividing) ✓

⚠ Mistake 4 — Differentiating the Right-Hand Side of a Constant as Zero, Incorrectly

When the equation is F(x, y) = 0 or F(x, y) = constant, differentiating the constant gives zero. But when the right-hand side is not a constant (it may be a function of x) it must be differentiated too. Always differentiate both sides completely.

§ 06Worked Exercises — 12 Problems

Work each problem before revealing the solution. The steps are written out in full, every chain rule application, every product rule, every algebraic rearrangement shown explicitly.

Exercises completed: 0 of 12

Group A — Polynomial Relations (Exercises 1 – 3)

Exercise 1 — Circle

Find dy/dx for x² + y² = 16

This is a circle of radius 4 centred at the origin. y cannot be isolated as a single function of x over the whole curve, so we differentiate implicitly.
Differentiate both sides with respect to x. Treat each term separately: d/dx[x²] + d/dx[y²] = d/dx[16]
Compute each derivative: d/dx[x²] = 2x d/dx[y²] = 2y · (dy/dx) (chain rule on the y² term) d/dx[16] = 0 (constant)
Assemble: 2x + 2y · (dy/dx) = 0
Isolate dy/dx, subtract 2x from both sides: 2y · (dy/dx) = −2x
Divide both sides by 2y (assuming y ≠ 0): dy/dx = −2x ÷ 2y = −x ÷ y The answer is in terms of both x and y. To find the gradient at a specific point, substitute both coordinates. For example at (2, 2√3): dy/dx = −2/(2√3) = −1/√3. Answer: dy/dx = −x ÷ y

Exercise 2 — Ellipse

Find dy/dx for 4x² + 9y² = 36

This is an ellipse. Differentiate both sides with respect to x: d/dx[4x²] + d/dx[9y²] = d/dx[36]
Compute each term: d/dx[4x²] = 8x d/dx[9y²] = 9 · 2y · (dy/dx) = 18y · (dy/dx) (chain rule) d/dx[36] = 0
Assemble the differentiated equation: 8x + 18y · (dy/dx) = 0
Solve for dy/dx: 18y · (dy/dx) = −8x dy/dx = −8x ÷ 18y = −4x ÷ 9y Notice how the coefficients 4 and 9 from the original ellipse equation appear in the denominator, a pattern that generalises: for the ellipse ax²+by²=c, dy/dx = −ax/(by). Answer: dy/dx = −4x ÷ 9y

Exercise 3 — Cubic Relation with Mixed Term

Find dy/dx for x³ + y³ = 6xy

This is the folium of Descartes. The right-hand side contains xy (a product of both variables) so we will need the product rule there. Differentiate both sides with respect to x: d/dx[x³] + d/dx[y³] = d/dx[6xy]
Left-hand side: d/dx[x³] = 3x² d/dx[y³] = 3y² · (dy/dx) (chain rule)
Right-hand side, use the product rule on 6xy: d/dx[6xy] = 6 · [d/dx[x] · y + x · d/dx[y]] = 6[1 · y + x · (dy/dx)] = 6y + 6x(dy/dx)
Assemble: 3x² + 3y²(dy/dx) = 6y + 6x(dy/dx)
Collect all dy/dx terms on the left: 3y²(dy/dx) − 6x(dy/dx) = 6y − 3x²
Factorise dy/dx on the left: (dy/dx)(3y² − 6x) = 6y − 3x²
Divide both sides by (3y² − 6x). Factor 3 from both sides: dy/dx = (6y − 3x²) ÷ (3y² − 6x) = 3(2y − x²) ÷ 3(y² − 2x) = (2y − x²) ÷ (y² − 2x) Answer: dy/dx = (2y − x²) ÷ (y² − 2x)

Group B — Trigonometric Relations (Exercises 4 – 6)

Exercise 4 — sin y in terms of x

Find dy/dx for sin y = x

This equation defines y implicitly as the inverse sine of x. Differentiating implicitly will give us d/dx[arcsin x] from first principles. This is genuinely useful: we are deriving a standard result without needing to know the answer in advance.
Differentiate both sides with respect to x: d/dx[sin y] = d/dx[x]
Left side, sin y is a composite function (sin of y, where y is a function of x). Chain rule applies: cos y · (dy/dx) = 1
Solve for dy/dx: dy/dx = 1 ÷ cos y
Express in terms of x. Since sin y = x, we need cos y in terms of x. Using sin²y + cos²y = 1: cos²y = 1 − sin²y = 1 − x² cos y = √(1 − x²) (taking positive root for y ∈ (−π/2, π/2))
Substitute: dy/dx = 1 ÷ √(1 − x²) This is the standard result d/dx[arcsin x] = 1/√(1−x²), now fully derived using implicit differentiation. See the Inverse Trig Derivatives page for all six inverse trig derivatives derived this way. Answer: dy/dx = 1 ÷ √(1 − x²)

Exercise 5 — Mixed Trig Relation

Find dy/dx for sin(x + y) = x

The argument of the sine is x + y, a function of both x and y. When we differentiate the left side with respect to x, the chain rule applies: The outer function is sin(·). The inner function is x + y. Since y depends on x, the inner derivative is d/dx[x + y] = 1 + dy/dx.
Differentiate both sides: d/dx[sin(x + y)] = d/dx[x]
Apply the chain rule on the left: cos(x + y) · d/dx[x + y] = 1 cos(x + y) · (1 + dy/dx) = 1
Expand the left side: cos(x + y) + cos(x + y) · (dy/dx) = 1
Isolate the dy/dx term: cos(x + y) · (dy/dx) = 1 − cos(x + y)
Divide by cos(x + y): dy/dx = [1 − cos(x + y)] ÷ cos(x + y) = 1/cos(x+y) − 1 (optional alternative form) Answer: dy/dx = [1 − cos(x + y)] ÷ cos(x + y)

Exercise 6 — x·cos y = y

Find dy/dx for x cos y = y

The left side x cos y is a product of x and cos y, both change with x. Product rule needed there, plus chain rule on cos y.
Differentiate both sides with respect to x: d/dx[x cos y] = d/dx[y]
Left side: product rule with u = x (u′ = 1) and v = cos y (v′ = −sin y · dy/dx by chain rule): d/dx[x cos y] = 1 · cos y + x · (−sin y · dy/dx) = cos y − x sin y · (dy/dx)
Right side: d/dx[y] = dy/dx
Assemble: cos y − x sin y · (dy/dx) = dy/dx
Collect dy/dx terms on the right (or left: choice): cos y = dy/dx + x sin y · (dy/dx) cos y = dy/dx(1 + x sin y)
Divide: Answer: dy/dx = cos y ÷ (1 + x sin y)

Group C — Exponential & Logarithm Relations (Exercises 7 – 8)

Exercise 7 — eˣʸ = x + y

Find dy/dx for e^(xy) = x + y

The left side is e raised to the power xy. Both x and y are functions of x (y is a function of x). To differentiate e^(xy), the chain rule applies: outer function is e^(·), inner function is xy.
To differentiate the inner function xy with respect to x, use the product rule: d/dx[xy] = y + x(dy/dx)
Differentiate both sides of the equation: d/dx[e^(xy)] = d/dx[x + y]
Left side: chain rule: e^(xy) · d/dx[xy] = e^(xy) · (y + x · dy/dx)
Right side: d/dx[x + y] = 1 + dy/dx
Assemble: e^(xy)(y + x · dy/dx) = 1 + dy/dx
Expand the left side: y · e^(xy) + x · e^(xy) · (dy/dx) = 1 + dy/dx
Collect all dy/dx terms on the left: x · e^(xy) · (dy/dx) − dy/dx = 1 − y · e^(xy) (dy/dx)(x · e^(xy) − 1) = 1 − y · e^(xy)
Divide: Answer: dy/dx = (1 − y · e^(xy)) ÷ (x · e^(xy) − 1)

Exercise 8 — ln(x + y) = y²

Find dy/dx for ln(x + y) = y²

Left side: ln(x + y) (outer function is ln(·), inner function is x + y. Right side: y²) chain rule applies.
Differentiate both sides: d/dx[ln(x + y)] = d/dx[y²]
Left side: chain rule. Derivative of ln(u) is 1/u, and d/dx[x + y] = 1 + dy/dx: [1/(x + y)] · (1 + dy/dx)
Right side, chain rule. Derivative of y² is 2y · dy/dx: 2y · dy/dx
Assemble: (1 + dy/dx) ÷ (x + y) = 2y · (dy/dx)
Multiply both sides by (x + y) to clear the denominator: 1 + dy/dx = 2y(x + y) · (dy/dx)
Collect dy/dx terms: 1 = 2y(x + y) · (dy/dx) − dy/dx 1 = (dy/dx)[2y(x + y) − 1]
Divide: Answer: dy/dx = 1 ÷ [2y(x + y) − 1]

Group D — Tangent Lines & Second Derivatives (Exercises 9 – 12)

Exercise 9 — Tangent Line to an Ellipse

Find the equation of the tangent to x² + 3y² = 7 at the point (2, 1)

First, verify the point (2, 1) lies on the curve: 2² + 3(1²) = 4 + 3 = 7 ✓
Find dy/dx by implicit differentiation: d/dx[x² + 3y²] = d/dx[7] 2x + 6y · (dy/dx) = 0
Solve for dy/dx: dy/dx = −2x ÷ 6y = −x ÷ 3y
Evaluate the gradient at (2, 1) by substituting x = 2, y = 1: dy/dx = −2 ÷ (3 · 1) = −2/3
Use point-slope form of a line: y − y₁ = m(x − x₁) with m = −2/3, x₁ = 2, y₁ = 1: y − 1 = −(2/3)(x − 2)
Expand and rearrange: y = 1 − (2/3)x + 4/3 y = −(2/3)x + 7/3 Or in the form: 2x + 3y = 7. This is a clean alternative form worth checking: 2(2)+3(1) = 7 ✓ Answer: y = −(2/3)x + 7/3 (or 2x + 3y = 7)

Exercise 10 — Tangent to a Curve Defined by a Cubic

Find the gradient of x³ + y³ = 2 at the point (1, 1)

Verify (1,1) lies on the curve: 1³ + 1³ = 2 ✓
Differentiate both sides with respect to x: 3x² + 3y² · (dy/dx) = 0
Solve for dy/dx: 3y²(dy/dx) = −3x² dy/dx = −x² ÷ y²
Evaluate at (1, 1): dy/dx = −1² ÷ 1² = −1 The curve has gradient −1 at this point. The tangent has a negative slope of 1, so it falls at 45° as you move right. Answer: gradient = −1 at (1, 1)

Exercise 11 — Second Derivative Implicitly

Find d²y/dx² for x² + y² = r² (the circle of radius r)

First, find dy/dx by implicit differentiation (we have done this for the unit circle, the result is the same for any radius r): 2x + 2y(dy/dx) = 0 → dy/dx = −x/y
Now differentiate dy/dx with respect to x again to find d²y/dx². We need d/dx[−x/y]. Treat this as a quotient: u = −x (u′ = −1), v = y (v′ = dy/dx): d²y/dx² = d/dx[−x/y] = (−1 · y − (−x) · dy/dx) ÷ y² = (−y + x · dy/dx) ÷ y²
Substitute dy/dx = −x/y: = (−y + x · (−x/y)) ÷ y² = (−y − x²/y) ÷ y²
Multiply numerator and denominator by y to clear the nested fraction: = (−y² − x²) ÷ y³
Recognise that x² + y² = r²: = −(x² + y²) ÷ y³ = −r² ÷ y³ The second derivative is negative everywhere on the top half of the circle (where y > 0) and positive on the bottom half (y < 0). This tells us the circle is concave down above the x-axis and concave up below: exactly as expected geometrically. Answer: d²y/dx² = −r² ÷ y³ (= −(x² + y²) ÷ y³)

Exercise 12 — Normal Line to a Hyperbola

Find the equation of the normal to xy = 4 at the point (2, 2)

Verify (2, 2) lies on xy = 4: 2 × 2 = 4 ✓ Note: this could also be solved by writing y = 4/x explicitly, but we demonstrate the implicit method here.
Differentiate both sides of xy = 4 with respect to x. The left side is a product: use the product rule: d/dx[xy] = d/dx[4] y + x(dy/dx) = 0
Solve for dy/dx: x(dy/dx) = −y dy/dx = −y/x
Evaluate the tangent gradient at (2, 2): dy/dx = −2/2 = −1
The normal is perpendicular to the tangent. If the tangent gradient is m, the normal gradient is −1/m: normal gradient = −1/(−1) = 1
Use point-slope form with gradient 1 and point (2, 2): y − 2 = 1 · (x − 2) y = x The normal to the rectangular hyperbola xy = 4 at (2, 2) is the line y = x. By symmetry of the hyperbola about y = x, this makes geometric sense. Answer: normal line is y = x

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