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Implicit Differentiation Calculator – Find dy/dx Step-by-Step

Implicit Differentiation Calculator — Find dy/dx Step-by-Step
Calculus Tool

Implicit Differentiation Calculator

Find dy/dx for any implicit equation F(x,y) = 0 with full term-by-term working — every y-term annotated with its chain-rule dy/dx factor. Evaluates dy/dx at a point, finds tangent line equations, and computes second derivatives d²y/dx².

Implicit Differentiation Solver — Find dy/dx
f(x,y) = 0 → dy/dx = ?
x^2 + y^2= 25 (circle)
x^3 + y^3= 6*x*y
x*y= 1
sin(y)= x
e^y= x^2 + y
x^2*y^3 + x*y= 10
ln(x*y)= x + y
x^2 - x*y + y^2= 4
cos(x+y)= x*y

Enter LHS and RHS separately. Use * for multiplication, ^ for powers. y is the dependent variable. Do NOT write dy/dx in the input — the calculator finds it for you.

Left side (LHS):
=
Right side (RHS):
x²+y²=25
x³+y³=6xy
xy=1
sin(y)=x
x²−xy+y²=4
eʸ=x²+y
ln(xy)=x+y
x²y³+xy=10
Error

Implicit Derivative — dy/dx

dy/dx = —
Step-by-Step Implicit Differentiation — Term-by-Term Working
Evaluate dy/dx at a Point (x₀, y₀) — Optional
x₀ =
y₀ =
Tangent Line Equation
—

Second Derivative via Implicit Differentiation: First find dy/dx, then differentiate dy/dx again with respect to x — treating y as a function of x throughout. Use quotient rule + chain rule, then substitute dy/dx back in.

Left side (LHS):
=
Right side (RHS):
x²+y²=25
x³+y³=6xy
xy=1
x²−xy+y²=4
Error
Implicit Differentiation — Common Cases Reference
1 Circle & Conic Sections x²+y²=r² → dy/dx = −x/y

For x²+y²=r²: differentiate both sides term-by-term. d/dx[x²] = 2x. d/dx[y²] = 2y·(dy/dx) — chain rule applies since y = y(x). d/dx[r²] = 0. So 2x + 2y·(dy/dx) = 0 → dy/dx = −x/y.

This is why the tangent to a circle at (x₀,y₀) is perpendicular to the radius — slope of radius = y₀/x₀, slope of tangent = −x₀/y₀, product = −1 ✓

2 Product of x and y Terms d/dx[xy] = x·dy/dx + y

When x and y are multiplied, use the product rule AND the chain rule together. d/dx[xy] = x·d/dx[y] + y·d/dx[x] = x·(dy/dx) + y. This is the most commonly missed step.

For xy = 1: (x·dy/dx + y) = 0 → x·dy/dx = −y → dy/dx = −y/x

3 Trigonometric Implicit sin(y)=x → dy/dx=1/cos(y)

For sin(y) = x: differentiate both sides. d/dx[sin(y)] = cos(y)·(dy/dx) — chain rule on y. d/dx[x] = 1. So cos(y)·(dy/dx) = 1 → dy/dx = 1/cos(y).

This is the proof that d/dx[arcsin(x)] = 1/√(1−x²) — since cos(y) = √(1−sin²y) = √(1−x²) ✓

4 Exponential Implicit e^y=x²+y → dy/dx=2x/(e^y−1)

For e^y = x² + y: d/dx[e^y] = e^y·(dy/dx). d/dx[x²+y] = 2x + dy/dx. So e^y·(dy/dx) = 2x + dy/dx → (e^y − 1)·(dy/dx) = 2x → dy/dx = 2x/(e^y − 1).

5 Folium of Descartes x³+y³=6xy → dy/dx=(2y−x²)/(y²−2x)

d/dx[x³+y³] = 3x² + 3y²·(dy/dx). d/dx[6xy] = 6(x·dy/dx + y) = 6x·(dy/dx) + 6y.

So 3x² + 3y²·(dy/dx) = 6x·(dy/dx) + 6y → (3y²−6x)·(dy/dx) = 6y−3x² → dy/dx = (6y−3x²)/(3y²−6x) = (2y−x²)/(y²−2x)

Implicit Differentiation — Reference Table

Click any row to load into the implicit differentiation calculator instantly.

Algebraic Curves
Equationdy/dxType
x²+y²=25−x/yCircle
xy=1−y/xHyperbola
x³+y³=6xy(2y−x²)/(y²−2x)Folium
x²−xy+y²=4(2x−y)/(x−2y)Ellipse
x²y³+xy=10via product ruleMixed
Transcendental
Equationdy/dxType
sin(y)=x1/cos(y)Trig
eʸ=x²+y2x/(eʸ−1)Exponential
ln(xy)=x+y(y−xy)/(xy−x)Log
cos(x+y)=xyvia chain ruleTrig

What Is Implicit Differentiation?

This implicit differentiation calculator finds dy/dx for any implicit equation F(x,y) = 0 by differentiating both sides term-by-term with respect to x, annotating every y-term with its chain-rule dy/dx factor, then isolating dy/dx algebraically. It also evaluates dy/dx at specific points and finds tangent line equations — everything a student needs when working with curves that cannot be expressed as explicit functions y = f(x).

Most calculus problems involve explicit functions — equations already solved for y, like y = x² + 3x. Differentiation is straightforward: dy/dx = 2x + 3. But many important curves cannot be written in explicit form globally. Consider x² + y² = 25 (a circle). Solving for y gives y = ±√(25 − x²) — two branches. Differentiating either branch is messy and misses the full picture.

Implicit differentiation avoids this problem entirely. You differentiate both sides of x² + y² = 25 directly with respect to x, treating y as an unknown function of x. The result is dy/dx = −x/y — a clean formula that works anywhere on the circle (except at y = 0, where the tangent is vertical). This is the power of implicit differentiation: dy/dx exists and is computable even when y cannot be isolated.

d/dx[F(x,y)] = 0  →  solve for dy/dx Equivalently: dy/dx = −Fx/Fy (Implicit Function Theorem)

How to Find dy/dx Implicitly — Step-by-Step Method

The implicit differentiation solver applies this four-step method for every equation. Mastering these steps allows you to find dy/dx for any implicit relation.

  1. Step 1 — Differentiate both sides with respect to x: Write d/dx[LHS] = d/dx[RHS] and apply it to every term.
  2. Step 2 — Every y-term picks up a dy/dx factor: Because y is a function of x, differentiating any expression involving y requires the chain rule — every y-term must be multiplied by dy/dx. For example, d/dx[y²] = 2y·(dy/dx), not 2y.
  3. Step 3 — Collect all dy/dx terms on one side: Move every term containing dy/dx to the left and everything else to the right.
  4. Step 4 — Factor out dy/dx and solve: dy/dx = (right side) / (coefficient of dy/dx on left).

Example 1: x² + y² = 25 (Circle)

  1. Differentiate both sides: d/dx[x²] + d/dx[y²] = d/dx[25]
  2. Term by term: 2x + 2y·(dy/dx) = 0 ← y² picks up dy/dx
  3. Collect: 2y·(dy/dx) = −2x
  4. dy/dx = −x/y

Example 2: xy = 1 (Hyperbola) — Product Rule Required

  1. d/dx[xy] = d/dx[1]
  2. Product rule: x·(dy/dx) + y·1 = 0 ← both x and y present, need product rule
  3. x·(dy/dx) = −y
  4. dy/dx = −y/x

Example 3: x³ + y³ = 6xy (Folium of Descartes)

  1. d/dx[x³] + d/dx[y³] = d/dx[6xy]
  2. 3x² + 3y²·(dy/dx) = 6·[x·(dy/dx) + y] ← product rule on right
  3. 3x² + 3y²·(dy/dx) = 6x·(dy/dx) + 6y
  4. Collect dy/dx: (3y² − 6x)·(dy/dx) = 6y − 3x²
  5. dy/dx = (2y − x²)/(y² − 2x)

Example 4: e^y = x² + y

  1. d/dx[e^y] = d/dx[x²] + d/dx[y]
  2. e^y·(dy/dx) = 2x + dy/dx ← e^y picks up dy/dx via chain rule
  3. (e^y − 1)·(dy/dx) = 2x
  4. dy/dx = 2x/(e^y − 1)

Example 5: sin(y) = x

  1. d/dx[sin(y)] = d/dx[x]
  2. cos(y)·(dy/dx) = 1
  3. dy/dx = 1/cos(y) = sec(y)
  4. Note: Since sin(y) = x, cos(y) = √(1−x²), giving the well-known dy/dx = 1/√(1−x²)

Implicit Differentiation and the Chain Rule

Implicit differentiation is the chain rule applied to y. This is the key insight that unifies the technique with everything else in calculus.

When we treat y as a function of x (i.e., y = y(x)), differentiating any expression f(y) with respect to x requires the chain rule:

d/dx[f(y)] = f'(y) · dy/dx Chain rule: outer derivative f'(y) times inner derivative dy/dx

Every implicit differentiation step where a y-term picks up a dy/dx factor is an application of the chain rule formula f'(g(x))·g'(x), with g(x) = y(x) as the inner function and dy/dx = y'(x) as the inner derivative. Examples:

Expressiond/dx (implicit)Chain Rule Reasoning
y²2y · dy/dxf(u)=u², f'(u)=2u, u=y → 2y · dy/dx
y³3y² · dy/dxf(u)=u³, f'(u)=3u², u=y → 3y² · dy/dx
sin(y)cos(y) · dy/dxf(u)=sin(u), f'(u)=cos(u) → cos(y) · dy/dx
e^ye^y · dy/dxf(u)=eᵘ, f'(u)=eᵘ → e^y · dy/dx
ln(y)(1/y) · dy/dxf(u)=ln(u), f'(u)=1/u → (1/y) · dy/dx
xy (product)x · dy/dx + yProduct rule: x·(dy/dx) + y·1

For a complete reference on the chain rule and f'(g(x))·g'(x), see our ⛓️ Chain Rule Calculator.

Finding Tangent Lines Using Implicit Differentiation

One of the most important applications of implicit differentiation is finding tangent lines to curves that cannot be expressed as single-valued functions. The process uses dy/dx as the slope at a specific point.

Worked Example: Tangent to x² + y² = 25 at (3, 4)

  1. Verify the point is on the curve: 3² + 4² = 9 + 16 = 25 ✓
  2. Find dy/dx via implicit differentiation: dy/dx = −x/y
  3. Evaluate at (3,4): dy/dx|(₃,₄) = −3/4 (slope of tangent)
  4. Point-slope form: y − 4 = −(3/4)(x − 3)
  5. Simplify: y = −(3/4)x + 9/4 + 4 = −(3/4)x + 25/4
  6. Tangent line: y = −(3/4)x + 25/4

Notice that the tangent slope dy/dx = −3/4 and the radius slope = 4/3 are negative reciprocals — their product is −1, confirming the tangent is perpendicular to the radius, as expected for a circle.

Why implicit differentiation is necessary for tangent lines to circles: If you tried to write y = √(25−x²) (the upper semicircle only) and differentiated explicitly, you'd get dy/dx = −x/√(25−x²). Substituting x=3: −3/4. Same answer — but the implicit approach works on both semicircles simultaneously and requires no branch selection. For curves without clean explicit forms (folium, cassini ovals), implicit differentiation is the only viable approach.

Second Derivatives via Implicit Differentiation

After finding dy/dx implicitly, you can find the second derivative d²y/dx² by differentiating dy/dx once more with respect to x — again treating y as a function of x throughout.

Worked Example: d²y/dx² for x² + y² = 25

  1. From implicit differentiation: dy/dx = −x/y
  2. Differentiate dy/dx with respect to x using the quotient rule:
    d²y/dx² = d/dx[−x/y] = [−1·y − (−x)·(dy/dx)] / y²
  3. = [−y + x·(dy/dx)] / y²
  4. Substitute dy/dx = −x/y:
    = [−y + x·(−x/y)] / y² = [−y − x²/y] / y²
  5. = [(−y² − x²)/y] / y² = −(x²+y²)/y³
  6. Use the original equation x²+y² = 25:
    d²y/dx² = −25/y³

This result shows the circle has everywhere-negative second derivative for y > 0 (the upper semicircle is concave down) and everywhere-positive for y < 0 (concave up) — consistent with a circle's geometry.

Common Mistakes With Implicit Differentiation

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

  • ❌ Wrong: d/dx[y²] = 2y
  • ✅ Correct: d/dx[y²] = 2y · (dy/dx) — chain rule required since y = y(x)
  • Every single y-term must pick up a dy/dx factor. No exceptions.

Mistake 2 — Forgetting the Product Rule for xy Terms

  • ❌ Wrong: d/dx[xy] = dy/dx   or   d/dx[xy] = y
  • ✅ Correct: d/dx[xy] = x·(dy/dx) + y — product rule gives two terms
  • When x and y are multiplied together, both the x-part and y-part contribute to the derivative.

Mistake 3 — Sign Errors When Isolating dy/dx

  • ❌ Wrong for x²+y²=25: 2x + 2y·(dy/dx) = 0 → dy/dx = 2x/(2y) = x/y
  • ✅ Correct: 2y·(dy/dx) = −2x → dy/dx = −x/y (the −2x comes from moving 2x to the right)
  • Always move ALL non-dy/dx terms to one side before dividing.

Mistake 4 — Not Differentiating Constants to Zero

  • ❌ Wrong: x² + y² = 25 → 2x + 2y·(dy/dx) = 25
  • ✅ Correct: d/dx[25] = 0 → 2x + 2y·(dy/dx) = 0
  • The derivative of any constant is 0 — the right side vanishes for F(x,y)=constant equations.

Mistake 5 — Treating dy/dx as a Fraction to Cancel Before Factoring

  • ❌ Wrong approach: 3y²·(dy/dx) − 6x·(dy/dx) = 6y − 3x² → "cancel" dy/dx → 3y² − 6x = (6y−3x²)/1 (incorrect!)
  • ✅ Correct: Factor dy/dx first → (3y²−6x)·(dy/dx) = 6y−3x² → then divide → dy/dx = (6y−3x²)/(3y²−6x)
  • You must collect all dy/dx terms and factor before dividing.

Worked Examples — Ten Full Implicit Differentiation Solutions

Equationdy/dxKey Rule Used
x²+y²=25−x/yPower rule + chain rule on y²
x³+y³=6xy(2y−x²)/(y²−2x)Power + product rule
xy=1−y/xProduct rule
sin(y)=x1/cos(y)Chain rule on sin(y)
x²−xy+y²=4(2x−y)/(x−2y)Product rule on xy
eʸ=x²+y2x/(eʸ−1)Chain rule on eʸ
ln(xy)=x+y(y−xy)/(xy−x)Log + product rule
x²y³+xy=10−(2xy³+y)/(3x²y²+x)Product rule (twice)
cos(x+y)=xyvia chain ruleChain + product rule
x²+y²=25 (2nd deriv.)d²y/dx²=−25/y³Quotient rule on −x/y

Full Solution 6: ln(xy) = x + y

  1. Use ln(xy) = ln(x) + ln(y): d/dx[ln(x)+ln(y)] = d/dx[x+y]
  2. 1/x + (1/y)·(dy/dx) = 1 + dy/dx
  3. Collect dy/dx: (1/y − 1)·(dy/dx) = 1 − 1/x = (x−1)/x
  4. (1−y)/y · (dy/dx) = (x−1)/x
  5. dy/dx = y(x−1)/[x(1−y)] = (xy−y)/(x−xy)
  6. dy/dx = (y−xy)/(xy−x) ✓

Full Solution 7: x²y³ + xy = 10

  1. d/dx[x²y³] + d/dx[xy] = d/dx[10]
  2. Product rule on x²y³: x²·3y²·(dy/dx) + 2x·y³
  3. Product rule on xy: x·(dy/dx) + y
  4. d/dx[10] = 0
  5. 3x²y²·(dy/dx) + 2xy³ + x·(dy/dx) + y = 0
  6. dy/dx·(3x²y²+x) = −2xy³ − y
  7. dy/dx = −(2xy³+y)/(3x²y²+x)

Frequently Asked Questions — Implicit Differentiation

What is implicit differentiation?
Implicit differentiation is a technique for finding dy/dx when y cannot be easily isolated as an explicit function of x. You differentiate both sides of the equation with respect to x, treating y as an implicit function of x. Every y-term picks up a dy/dx factor via the chain rule. Then you solve algebraically for dy/dx. The implicit differentiation calculator automates this and shows every step.
When do you need implicit differentiation instead of regular differentiation?
Use implicit differentiation when the equation defines y implicitly and you cannot (or it's impractical to) solve for y = f(x). Examples: x²+y²=25 (circle, two branches), x³+y³=6xy (folium), e^y=x²+y (y appears in exponent), ln(xy)=x+y. Any time the equation mixes x and y in ways that prevent clean isolation of y, the implicit differentiation solver is the right tool.
How does the chain rule apply to implicit differentiation?
Implicit differentiation IS the chain rule applied to y(x). Since y is an unknown function of x, d/dx[f(y)] = f'(y)·dy/dx — exactly the chain rule formula f'(g(x))·g'(x) with g(x)=y(x). So d/dx[y²]=2y·dy/dx, d/dx[sin(y)]=cos(y)·dy/dx, d/dx[e^y]=e^y·dy/dx. The dy/dx factor appears every time because y is the inner function.
How do you find the tangent line to an implicit curve?
Step 1: Find dy/dx using implicit differentiation. Step 2: Substitute (x₀, y₀) into dy/dx to get the slope m. Step 3: Write the tangent line as y − y₀ = m(x − x₀). For x²+y²=25 at (3,4): dy/dx = −x/y = −3/4, so tangent is y−4 = −(3/4)(x−3), giving y = −(3/4)x + 25/4. The implicit differentiation calculator does all of this automatically when you enter x₀ and y₀.
What is the implicit function theorem?
The implicit function theorem states that if F(x,y)=0 and ∂F/∂y ≠ 0 at a point, then y is locally a function of x near that point, and dy/dx = −(∂F/∂x)/(∂F/∂y) = −Fx/Fy. This is the mathematical foundation behind implicit differentiation — even when y cannot be isolated globally, its derivative dy/dx exists wherever Fy ≠ 0. Our implicit differentiation solver uses this as a computational shortcut internally, while displaying the term-by-term working for learning purposes.
Can you find d²y/dx² using implicit differentiation?
Yes — after finding dy/dx, differentiate the dy/dx expression again with respect to x, still treating y as a function of x. This uses the quotient rule and chain rule together, and you substitute dy/dx back in at the end. For x²+y²=25: dy/dx=−x/y, so d²y/dx² = [−y + x·dy/dx]/y² = [−y + x(−x/y)]/y² = −(x²+y²)/y³ = −25/y³. Our implicit differentiation calculator handles this in Tool 2.

Related Calculators

⛓️
⛓️ Chain Rule Calculator — the rule behind every dy/dx factor in implicit differentiation
∂
∂ Partial Derivative Calculator — Fx and Fy used in dy/dx = −Fx/Fy
∫
∫ Antiderivative Calculator — inverse of differentiation
Implicit Diff Quick Rules
d/dx[y] = dy/dx Base case — chain rule
d/dx[y²] = 2y·dy/dx Power + chain rule
d/dx[yⁿ] = n·yⁿ⁻¹·dy/dx General power
d/dx[xy] = x·dy/dx + y Product rule!
d/dx[sin(y)] = cos(y)·dy/dx Trig + chain rule
d/dx[e^y] = e^y·dy/dx Exp + chain rule
d/dx[ln(y)] = (1/y)·dy/dx Log + chain rule
dy/dx = −Fx/Fy Implicit Function Theorem
Most Searched
x²+y²=25 → dy/dx=−x/y
xy=1 → dy/dx=−y/x
x³+y³=6xy (Folium)
sin(y)=x → dy/dx=1/cos(y)
eʸ=x²+y → dy/dx=2x/(eʸ−1)
x²−xy+y²=4
ln(xy)=x+y
x²y³+xy=10
Related Tools
⛓️
Chain Rule Calculator
∂
Partial Derivative Calc
∫
Antiderivative Calculator

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