Learning Objectives
Differentiate a relation while treating y as a function of x locally.
Solve for dy/dx after applying chain and product rules.
Find a tangent slope at a point and recognize where the derivative formula fails.
Prerequisites: the chain rule, product rule, and tangent-line equations. Implicit differentiation is useful when x and y are linked by an equation that is inconvenient to solve explicitly for y.
1. Differentiate Both Sides with Respect to x
When y depends on x, differentiating y² requires the chain rule: d(y²)/dx = 2y(dy/dx). Write y′ for dy/dx if convenient. In contrast, d(x²)/dx = 2x because the derivative of x with respect to itself is 1.
Worked Example: A Circle
Differentiate x² + y² = 25. The result is 2x + 2yy′ = 0.
Solve for y′: 2yy′ = −2x, so y′ = −x/y wherever y ≠ 0 on a differentiable local branch.
At the point (3, 4), the slope is −3/4. The tangent line is y − 4 = (−3/4)(x − 3). The point satisfies the original relation because 9 + 16 = 25.
2. A Product Involving y Needs Both Rules
For the product xy, both factors depend on x: the first directly, the second through y(x). Therefore d(xy)/dx = y + xy′. Differentiating it as just y or just xy′ misses one term.
Worked Example: Collect the Derivative Terms
For x² + xy + y² = 7, differentiate to obtain 2x + y + xy′ + 2yy′ = 0.
Collect terms containing y′: (x + 2y)y′ = −(2x + y). Thus y′ = −(2x + y)/(x + 2y), when x + 2y ≠ 0.
At (1, 2), the relation holds because 1 + 2 + 4 = 7. The slope is −4/5, so the tangent line is y − 2 = (−4/5)(x − 1).
3. Check That the Point Lies on the Curve
A derivative expression may accept a numerical pair that is not on the original relation, but the resulting number would not be a tangent slope to that curve at that point. Substitute the point into the original equation before reporting a tangent line.
4. Zero Denominators Need Geometric Analysis
For the circle x² + y² = 25, the formula −x/y fails when y = 0. At (5, 0) and (−5, 0), the tangent lines are vertical, x = 5 and x = −5, so there is no finite derivative dy/dx there.
A zero denominator in an implicit derivative does not always prove a vertical tangent by itself. If both numerator and denominator vanish, the relation may have a more complicated local shape. Check the original curve rather than dividing by zero or assigning a slope without justification.
5. Differentiate Again Carefully
Example: A Second Derivative on the Circle
Starting from y′ = −x/y, apply the quotient rule: y″ = [−y + xy′]/y².
Substitute y′ = −x/y to get y″ = −(x² + y²)/y³. On the circle, x² + y² = 25, so y″ = −25/y³ where y ≠ 0.
At (3, 4), the second derivative is −25/64. Even when a formula contains y, it is still differentiated as a function of x.
Common Mistakes
Attach y′ when differentiating a nonconstant function of y with respect to x.
Use the product rule for terms such as xy.
Solve for y′ only after collecting every term that contains it.
Check the point and any denominator restrictions before substitution.
6. Practice
1. Find dy/dx for x² + 4y² = 16.
2. Find the tangent line to that ellipse at (0, 2).
3. Find dy/dx for xy = 6 and evaluate it at (2, 3).
4. Differentiate sin y = x² implicitly on a differentiable branch where cos y ≠ 0.
Worked Solutions
1. Differentiation gives 2x + 8yy′ = 0, so y′ = −x/(4y) when y ≠ 0.
2. The point lies on the ellipse and the slope is 0. The tangent line is y = 2.
3. The product rule gives y + xy′ = 0, so y′ = −y/x for x ≠ 0. At (2, 3), the slope is −3/2.
4. The chain rule gives (cos y)y′ = 2x, so y′ = 2x/cos y under the stated condition.
7. Summary
- Treat y as depending on x during implicit differentiation.
- Combine chain and product rules, then isolate y′.
- A tangent calculation requires a point on the original relation.
- Analyze exceptional points separately when the solved formula has a zero denominator.