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Implicit Differentiation

Implicit Differentiation

A method for finding the derivative of an implicitly defined function or relation.

 

Example showing implicit differentiation to find dy/dx for x²+xy−y²=1, with step-by-step algebraic solution yielding...

 

 

See also

Derivative rules

Key Formula

ddx[f(y)]=f′(y)⋅dydx\frac{d}{dx}[f(y)] = f'(y) \cdot \frac{dy}{dx}
Where:
  • yy = A variable that depends on x, even though it is not written as an explicit function of x
  • f(y)f(y) = Any expression involving y
  • dydx\frac{dy}{dx} = The derivative of y with respect to x, which is the quantity you are solving for

Worked Example

Problem: Find dy/dx for the circle x² + y² = 25.
Step 1: Differentiate both sides of the equation with respect to x. For the x² term, you get 2x. For the y² term, apply the chain rule: the outer derivative gives 2y, and you multiply by dy/dx because y depends on x. The right side, 25, is a constant, so its derivative is 0.
ddx[x2]+ddx[y2]=ddx[25]\frac{d}{dx}[x^2] + \frac{d}{dx}[y^2] = \frac{d}{dx}[25]
Step 2: Write out the derivatives.
2x+2y dydx=02x + 2y\,\frac{dy}{dx} = 0
Step 3: Isolate dy/dx by subtracting 2x from both sides, then dividing by 2y.
2y dydx=−2x  ⟹  dydx=−2x2y=−xy2y\,\frac{dy}{dx} = -2x \implies \frac{dy}{dx} = \frac{-2x}{2y} = -\frac{x}{y}
Step 4: Check at a known point. The point (3, 4) lies on the circle because 9 + 16 = 25. The derivative there is −3/4, which gives a negative slope in the first quadrant — consistent with the circle curving downward.
dydx∣(3,4)=−34\frac{dy}{dx}\bigg|_{(3,4)} = -\frac{3}{4}
Answer: dy/dx = −x/y

Another Example

Problem: Find dy/dx for the equation x³ + y³ = 6xy.
Step 1: Differentiate every term with respect to x. On the left, x³ gives 3x² and y³ gives 3y²·(dy/dx) by the chain rule. On the right, 6xy requires the product rule.
3x2+3y2 dydx=6y+6x dydx3x^2 + 3y^2\,\frac{dy}{dx} = 6y + 6x\,\frac{dy}{dx}
Step 2: Collect all terms containing dy/dx on one side and everything else on the other.
3y2 dydx−6x dydx=6y−3x23y^2\,\frac{dy}{dx} - 6x\,\frac{dy}{dx} = 6y - 3x^2
Step 3: Factor out dy/dx and solve.
dydx(3y2−6x)=6y−3x2  ⟹  dydx=6y−3x23y2−6x=2y−x2y2−2x\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2 \implies \frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}
Answer: dy/dx = (2y − x²) / (y² − 2x)

Frequently Asked Questions

When do you use implicit differentiation instead of regular differentiation?
Use implicit differentiation whenever the equation relating x and y cannot be easily solved for y, or when solving for y would create multiple branches (like the top and bottom halves of a circle). It is also the standard method for equations like x² + y² = 25 or sin(xy) = x, where isolating y is impractical or impossible.
Why do you multiply by dy/dx when differentiating a y term?
Because y is treated as a function of x, not as an independent variable. Whenever you differentiate a function of y with respect to x, the chain rule requires you to multiply the outer derivative by the derivative of the inner function — that inner derivative is dy/dx. Forgetting this step is the single most common error in implicit differentiation.

Explicit Differentiation vs. Implicit Differentiation

With explicit differentiation, you start from a formula like y = f(x) and differentiate directly. With implicit differentiation, x and y are intertwined in one equation (e.g., x² + y² = 25), so you differentiate both sides with respect to x and solve for dy/dx. Explicit differentiation is a special case — if you can isolate y first, both methods give the same result.

Why It Matters

Many important curves and surfaces cannot be written as y = f(x). Circles, ellipses, hyperbolas, and the curves defined by equations like x³ + y³ = 6xy (the folium of Descartes) all require implicit differentiation to find slopes and tangent lines. In physics and engineering, relationships between variables are often given implicitly, making this technique essential for computing rates of change.

Common Mistakes

Mistake: Forgetting to attach dy/dx when differentiating terms containing y.
Correction: Every time you differentiate a function of y with respect to x, the chain rule produces a factor of dy/dx. For example, d/dx[y²] = 2y·(dy/dx), not just 2y.
Mistake: Forgetting to use the product rule or quotient rule on mixed terms like xy or x²y³.
Correction: A term like xy involves two functions of x (since y depends on x). Apply the product rule: d/dx[xy] = x·(dy/dx) + y·1. Skipping the product rule leads to missing terms and incorrect results.

Related Terms