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Implicit differentiation — checked on the curve itself

Type a relation like x² + y² = 25. You get dy/dx with every term differentiated in the open — and a verification performed at an actual point on your curve, not a hand-wave.

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  • ✓Every answer verified before you see it

In short

Implicit differentiation finds dy/dx when y is tangled inside a relation like x² + y² = 25 instead of standing alone as y = f(x). VerifySolve differentiates term by term, applying the chain rule to every y-term, solves for dy/dx, and checks the result numerically at a real point that lies on your curve.

The implicit differentiation moves

One idea runs the whole technique: y is a function of x, so every y-term is a chain-rule problem. These are the moves the solver makes, in order.

y-powers (chain rule)

ddx yn=nyn−1⋅dydx\dfrac{d}{dx}\, y^n = n y^{n-1} \cdot \dfrac{dy}{dx}

Products with y

ddx (xy)=y+x⋅dydx\dfrac{d}{dx}\, (xy) = y + x \cdot \dfrac{dy}{dx}

The strategy

Differentiate both sides  →  collect dydx terms  →  divide\text{Differentiate both sides} \;\rightarrow\; \text{collect } \tfrac{dy}{dx} \text{ terms} \;\rightarrow\; \text{divide}

The check identity

Fx+Fy⋅dydx=0F_x + F_y \cdot \dfrac{dy}{dx} = 0

Worked examples — verified at real points

Each verification below happened at an actual point satisfying the equation — the same check the live solver runs on your relation.

x² + y² = 25

  1. 1.Differentiate term by term: 2x + 2y·y′ = 0
  2. 2.Collect y′: 2y·y′ = −2x

dy/dx = −x/y

At (3, 4): slope = −3/4, and the curve identity checks out ✓. This is the circle of radius 5.

xy = 6

  1. 1.Product rule on xy: y + x·y′ = 0
  2. 2.Collect y′: x·y′ = −y

dy/dx = −y/x

Verified at a point on the hyperbola ✓.

y² = 4x

  1. 1.Differentiate: 2y·y′ = 4
  2. 2.Divide by 2y

dy/dx = 2/y

Verified at (1, −2) on the parabola ✓. Note the slope is never defined at the vertex (0, 0) — the tangent there is vertical.

x² + xy + y² = 7

  1. 1.Term by term: 2x + (y + x·y′) + 2y·y′ = 0
  2. 2.Collect y′ terms: y′(x + 2y) = −(2x + y)

dy/dx = −(2x + y)/(x + 2y)

Verified numerically at a found point on the curve ✓ — product terms and y² terms together.

Common mistakes — and how to dodge them

Differentiating y² as 2y and moving on

That is the derivative if y were the variable — but we are differentiating with respect to x. Every y-term needs its chain-rule factor: d/dx [y²] = 2y · y′. Missing it is the most common error in the whole topic.

Sign errors when collecting y′

Moving −2x across the equation flips its sign; moving the y′ coefficient means dividing, not subtracting. Write the “collect” step out in full — the solver does, and it is where most hand solutions quietly go wrong.

Plugging in a point that is not on the curve

dy/dx = −x/y gives a slope only where the relation actually holds. (3, 4) is on x² + y² = 25 (9 + 16 = 25 ✓); (3, 3) is not, and its “slope” would be meaningless. Check membership before substituting.

Implicit Differentiation Calculator — questions

When do you need implicit differentiation?

When y cannot be isolated — or isolating it would be a mess. Circles (x² + y² = 25), curves like x² + xy + y² = 7, and relations where y appears in several places are the classic cases. Instead of solving for y first, you differentiate both sides with y treated as a function of x, then collect the dy/dx terms.

How is an implicit derivative verified?

At a real point on the curve. The calculator finds an actual (x, y) pair satisfying your equation numerically, measures ∂F/∂x and ∂F/∂y there by finite differences on your original relation, and checks the identity Fx + Fy · dy/dx = 0 with our formula’s value. If the identity holds at a genuine point of the curve, the formula is consistent with the relation — and you see the numbers.

Why does every y-term grow a dy/dx factor?

The chain rule. y is secretly a function of x, so differentiating y² gives 2y · (dy/dx) — the derivative of the outside times the derivative of the inside. Forgetting that factor is the number-one implicit differentiation error.

How do I find the slope at a specific point?

Get the dy/dx formula from the calculator, then substitute the point. For x² + y² = 25, dy/dx = −x/y, so at (3, 4) the slope is −3/4. Check the point really is on the curve first (3² + 4² = 25 ✓).

What relations are covered?

Polynomial relations in x and y — powers of x and y, products like xy, and sums of them — which covers the standard calculus course cases. Relations with sin(y), e^y and similar are not covered yet; the calculator says so instead of guessing.

Is the implicit differentiation calculator free?

Yes — free, no sign-up, no app, with full steps and the point-on-the-curve verification included.

Keep solving

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The pre-calculus toolkit, all verified.

A note on honest use: VerifySolve is built for learning. Use the steps to understand the method, then close the tab and try one on your own — that is where the grade actually comes from. If your teacher or school has a policy on calculators, follow it.