The method for every problem on this page is the same: differentiate both sides with respect to x, treat every y as a function of x (so d/dx[y²] = 2y·y'), collect all the y' terms on one side, then factor and divide. The algebra varies; the method doesn't. Work each problem through before checking the solution.

The Method in 3 Steps

📋 Implicit Differentiation Method
Step 1Differentiate both sides with respect to x. When differentiating y, apply chain rule: d/dx[y] = dy/dx = y'
Step 2Collect all terms with y' (dy/dx) on one side
Step 3Factor out y' and divide to solve

Problem 1 — Circle x² + y² = 25

📋 Find dy/dx
Differentiate2x + 2y·y' = 0
Solvey' = −x/y

Problem 2 — x³ + y³ = 8

📋 Find dy/dx
Differentiate3x² + 3y²·y' = 0
Solvey' = −x²/y²

Problem 3 — x²y + xy² = 6 (Product Rule Required)

📋 Find dy/dx
d/dx[x²y]= 2xy + x²y' (product rule)
d/dx[xy²]= y² + 2xyy' (product rule)
Combine2xy + x²y' + y² + 2xyy' = 0
Collect y'y'(x² + 2xy) = −2xy − y²
Answery' = −(2xy + y²)/(x² + 2xy)

Problem 4 — sin(xy) = x

📋 Find dy/dx
Differentiatecos(xy)·(y + xy') = 1
Expandy·cos(xy) + xy'·cos(xy) = 1
Answery' = [1 − y·cos(xy)] / [x·cos(xy)]

Problem 5 — e^(xy) = x + y

📋 Find dy/dx
Differentiatee^(xy)·(y + xy') = 1 + y'
Expandye^(xy) + xye^(xy)·y' - y' = 1 (rearranging)
Answery' = (1 − ye^(xy)) / (xe^(xy) − 1)

Problem 6 — Tangent Line to x² + y² = 13 at (2, 3)

📋 Find the equation of the tangent line
Step 1From Problem 1: y' = −x/y
SlopeAt (2,3): slope = −2/3
Liney − 3 = −(2/3)(x − 2) → y = −(2/3)x + 13/3

Problem 7 — ln(y) = x² − y²

📋 Find dy/dx
Differentiatey'/y = 2x − 2yy'
Multiply by yy' = 2xy − 2y²y'
Collecty'(1 + 2y²) = 2xy
Answery' = 2xy / (1 + 2y²)

Problem 8 — Second Implicit Derivative of x² + y² = r²

📋 Find d²y/dx²
Firsty' = −x/y (from Problem 1)
Differentiate y'y'' = −(y − x·y')/y² = −(y − x·(−x/y))/y²
Simplifyy'' = −(y² + x²)/y³ = −r²/y³

Common Mistakes to Avoid

⚠ Most Common Errors

Forgetting the chain rule on y terms. d/dx[y²] = 2y·y', not just 2y. Every time you differentiate a y expression, you must multiply by y' because y is a function of x.

⚠ Product Rule

When a term contains both x and y (like x²y), you must use the product rule. d/dx[x²y] = 2xy + x²y'. Students frequently miss the x²y' term.

Frequently Asked Questions

When should I use implicit differentiation instead of solving for y?▾
Use implicit differentiation when solving for y is impossible or creates messy expressions. For example, x³ + y³ = 8 could be solved for y using a cube root, but implicit differentiation is much cleaner. Circles, ellipses, and any equation where y appears multiple times usually call for implicit differentiation.
Can I check my answer?▾
Yes — differentiate the original equation both ways. Solve for y explicitly if possible, differentiate directly, and compare. Or pick a specific point on the curve and verify that both formulas give the same slope.

Strategy for Complex Implicit Problems

For harder implicit differentiation problems — particularly those involving products of x and y, or trig/exponential terms — follow this approach: differentiate every term separately, apply the product rule to any xy product, collect every term containing y' on the left, and factor y' out. The algebra is usually straightforward once you have differentiated correctly.

Problems 9–12 — Advanced Level

📋 Problem 9 — x²y² + x³y = 10
Differentiate2xy² + x²·2y·y' + 3x²y + x³y' = 0
Collect y'y'(2x²y + x³) = −2xy² − 3x²y
Answery' = −xy(2y + 3x) / (x²(2y + x))
📋 Problem 10 — tan(xy) = y
Differentiatesec²(xy)·(y + xy') = y'
Expandy·sec²(xy) + xy'·sec²(xy) = y'
Collecty'(1 − x·sec²(xy)) = y·sec²(xy)
Answery' = y·sec²(xy) / (1 − x·sec²(xy))
📋 Problem 11 — Find all points on x² + y² = 25 with slope 3/4
y'= −x/y. Set −x/y = 3/4 → y = −4x/3
Substitutex² + 16x²/9 = 25 → 25x²/9 = 25 → x = ±3
Points(3, −4) and (−3, 4)
📋 Problem 12 — Second derivative of x² + y² = r²
Firsty' = −x/y
Secondy'' = −(y − xy')/y² = −(y − x(−x/y))/y² = −(y² + x²)/y³ = −r²/y³
AM
Dr. Aisha Malik, PhD Mathematics
Senior Lecturer in Applied Mathematics · 12 years teaching calculus
Dr. Malik holds a PhD in Applied Mathematics from the University of Edinburgh and has taught calculus to over 4,000 students. She has reviewed this article for mathematical accuracy and pedagogical clarity.
Technically reviewed by: Prof. James Chen, Stanford Mathematics Department · April 2026