Implicit Differentiation on a Circle: Remember the dy/dx Factor

Differentiate both sides with respect to x while treating y as a local function of x. For x² + y² = R², this gives 2x + 2y(dy/dx) = 0, so dy/dx = −x/y where y ≠ 0.

Why y² needs an extra factor

If y depends on x, differentiating y² uses the chain rule: its derivative is 2y times dy/dx. Writing only 2y would treat y as though it were the independent variable. Implicit differentiation is useful because it finds a slope without first isolating y using a square root.

A complete circle is not one function y of x, but near a point with y ≠ 0 its upper or lower branch can be treated as a differentiable function. Keep that local interpretation and the nonzero denominator condition in the final slope formula.

Worked example

Find the tangent slope to x² + y² = 169 at (5, 12). First confirm the point: 5² + 12² = 25 + 144 = 169. Differentiate the equation to obtain 2x + 2y y′ = 0. Rearranging gives y′ = −x/y.

At (5, 12), the slope is −5/12. The tangent line is y − 12 = −(5/12)(x − 5). As a geometric check, the radius from the origin to this point has slope 12/5. Multiplying the two finite slopes gives −1, consistent with a tangent perpendicular to the radius.

What happens when y = 0?

The formula −x/y is not a finite number at the circle’s left and right endpoints. The tangents there are vertical. You can see this geometrically or work locally with x as a function of y. Do not write a slope of zero simply because division is impossible.

At the top and bottom of the circle, x = 0 and y ≠ 0, so y′ = 0. Those tangents are horizontal. The same formula therefore distinguishes two different special cases rather than treating every axis point alike.

Other implicit products

For xy = 12, differentiate with the product rule: y + x y′ = 0. At (3, 4), y′ = −4/3. Both product rule and chain rule may appear in the same implicit problem. Gather the terms containing y′, factor it out, and then divide by its coefficient where permitted.

Common mistake

Always confirm that a supplied point lies on the curve before finding its slope. A derivative expression evaluated at an unrelated point does not give a tangent to the stated curve. Keep coordinates separate from the derivative until after differentiating, or you may turn variables into constants too early.

y bhi x ke saath badalta hai—y² ke derivative mein y′ mat bhoolo.

Three practice questions with worked answers

1. Find the slope of x² + y² = 100 at (6, 8).

   The point lies on the circle because 36 + 64 = 100. Use y′ = −x/y to obtain −6/8 = −3/4.

2. Find the slope of 4x² + y² = 25 at (2, 3).

   Differentiate: 8x + 2y y′ = 0, so y′ = −4x/y. At (2, 3), the slope is −8/3; the point satisfies 16 + 9 = 25.

3. Find the slope of xy = 12 at (3, 4).

   The product rule gives y + x y′ = 0. Substitute to get 4 + 3y′ = 0, hence y′ = −4/3.

Related study guides

• Circle Equation from Centre and Radius
• Differentiation Formulas Class 12: Complete List with Examples

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References and authorship

Original examples and practice questions, prepared with AI assistance and editorial arithmetic review. Related definitions and methods: OpenStax: Implicit Differentiation. No textbook or exam-board affiliation is claimed.