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Single Variable Calculus
Differentiation
01The Derivative as the Slope of the Tangent Line02Secant Lines and the Limit Definition of the Derivative03Differentiating 1/x Straight From the Definition04Tangents to 1/x and the Triangle of Area 205Newton and Leibniz Notation, and the Power RuleProblem set0/10Problem set 20/10MIT problem set0/3Practice∞
01The Derivative as an Instantaneous Rate of Change02Rates of Change Without Time: Gradient and Sensitivity03Easy Limits, 0/0, and One-Sided Limits04Continuity, Jumps, and Removable Discontinuities05Infinite Discontinuities: 1/x and Its Derivative06Differentiable Implies ContinuousProblem set0/10Problem set 20/10Practice∞
01Two Kinds of Formula, and the Derivative of Sine02The Derivative of Cosine, and Two Limits at Zero03Bow and Bowstring: Proving the Two Trig Limits04Why the Trig Derivatives Hold Only in Radians05A Geometric Proof: Sine as a Height on the Circle06Finishing the Proof, and the Product and Quotient RulesProblem set0/10Problem set 20/10MIT problem set0/1Practice∞
01Proving the Product Rule: Change One Factor at a Time02The Quotient Rule, and the Power Rule for Negative Exponents03The Chain Rule: Differentiating a Function of a Function04Higher Derivatives: The Sine Cycle and Three Notations05The nth Derivative of xⁿ Is n Factorial, by InductionProblem set0/10Problem set 20/10MIT problem set0/2Practice∞
01The Power Rule for Rational Exponents02The Slope of a Circle, Two Ways03Implicit Differentiation of a Quartic Curve04Inverse Functions and the Reflection Across y = x05Derivatives of the Arctangent and ArcsineProblem set0/10Problem set 20/10MIT problem set0/1Practice∞

Differentiating 1/x Straight From the Definition

Every difference quotient starts life as 0/0 — which cancellation rescues 1/x and lets the limit finally be taken?


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Your summary note

    1. 1

      The 0/00/00/0 pitfall and the cancellation

      Write the difference quotient for f(x)=1/xf(x)=1/xf(x)=1/x at x0x_0x0​, note that setting Δx=0\Delta x=0Δx=0 immediately gives 0/00/00/0, then combine over the common denominator (x0+Δx)x0(x_0+\Delta x)x_0(x0​+Δx)x0​ and factor 1/Δx1/\Delta x1/Δx out front.

    2. 2

      The result f′(x0)=−1/x02f'(x_0)=-1/x_0^2f′(x0​)=−1/x02​

      Carry the numerator to −Δx-\Delta x−Δx, cancel the Δx\Delta xΔx against the factor out front to reach −1/[(x0+Δx)x0]-1/[(x_0+\Delta x)x_0]−1/[(x0​+Δx)x0​], then set Δx=0\Delta x=0Δx=0 to get −1/x02-1/x_0^2−1/x02​.

    3. 3

      Graph sanity-check on the hyperbola

      Record that f′(x0)f'(x_0)f′(x0​) is negative everywhere matching the downward-sloping tangent, and that its magnitude shrinks as x0→∞x_0\to\inftyx0​→∞ matching the flattening curve.

    Attempt 1 of 2