Single Variable Calculus

Differentiation

01The Derivative as the Slope of the Tangent Line 02Secant Lines and the Limit Definition of the Derivative 03Differentiating 1/x Straight From the Definition 04Tangents to 1/x and the Triangle of Area 2 05Newton and Leibniz Notation, and the Power Rule Problem set0/10 Problem set 20/10 MIT problem set0/3 Practice∞

01The Derivative as an Instantaneous Rate of Change 02Rates of Change Without Time: Gradient and Sensitivity 03Easy Limits, 0/0, and One-Sided Limits 04Continuity, Jumps, and Removable Discontinuities 05Infinite Discontinuities: 1/x and Its Derivative 06Differentiable Implies Continuous Problem set0/10 Problem set 20/10 Practice∞

01Two Kinds of Formula, and the Derivative of Sine 02The Derivative of Cosine, and Two Limits at Zero 03Bow and Bowstring: Proving the Two Trig Limits 04Why the Trig Derivatives Hold Only in Radians 05A Geometric Proof: Sine as a Height on the Circle 06Finishing the Proof, and the Product and Quotient Rules Problem set0/10 Problem set 20/10 MIT problem set0/1 Practice∞

01Proving the Product Rule: Change One Factor at a Time 02The Quotient Rule, and the Power Rule for Negative Exponents 03The Chain Rule: Differentiating a Function of a Function 04Higher Derivatives: The Sine Cycle and Three Notations 05The nth Derivative of xⁿ Is n Factorial, by Induction Problem set0/10 Problem set 20/10 MIT problem set0/2 Practice∞

01The Power Rule for Rational Exponents 02The Slope of a Circle, Two Ways 03Implicit Differentiation of a Quartic Curve 04Inverse Functions and the Reflection Across y = x 05Derivatives of the Arctangent and Arcsine Problem set0/10 Problem set 20/10 MIT problem set0/1 Practice∞

The Slope of a Circle, Two Ways

Why does differentiating x^2 + y^2 = 1 as it stands beat solving for y, and how does y' = -x/y cover both halves of the circle at once?

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Your summary note

1. The explicit branch $y = (1 - x^2)^{1/2}$
  Rewrite the square root in fractional-power form, apply the chain rule together with the power rule for $a = \frac{1}{2}$ , and simplify to $y' = \frac{-x}{\sqrt{1 - x^2}}$ .
2. Implicit differentiation of $x^2 + y^2 = 1$
  Apply $\frac{d}{dx}$ to both sides as the equation stands, get $2x + 2y\,y' = 0$ with the chain rule on $y^2$ , solve for $y' = -\frac{x}{y}$ , and reconcile it with the explicit answer.
3. One formula for both halves of the circle
  Redo the comparison with $y = -\sqrt{1 - x^2}$ , tracking the extra minus signs on each side, then record the sign of $-\frac{x}{y}$ for positive $x$ on both halves.
Attempt 1 of 2