Ludium
Sign In
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∞

The Quotient Rule, and the Power Rule for Negative Exponents

Where does the quotient rule's minus sign come from, and how does setting the numerator to 1 stretch the power rule to negative exponents?


Loading…

←Previous Proving the Product Rule: Change One Factor at a TimeNext The Chain Rule: Differentiating a Function of a Function →

Your summary note

    1. 1

      The quotient rule (uv)′=u′v−uv′v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}(vu​)′=v2u′v−uv′​

      Write the formula down and record the order of the two terms, the minus sign, and the square of the old denominator as the asymmetry to watch.

    2. 2

      Derivation by the change-and-limit pattern

      Reproduce the derivation: form Δ(u/v)\Delta(u/v)Δ(u/v) over the common denominator (v+Δv)v(v+\Delta v)v(v+Δv)v, cancel the two copies of uvuvuv, divide by Δx\Delta xΔx, and take Δx→0\Delta x \to 0Δx→0 with vvv continuous.

    3. 3

      Reciprocal rule and ddxx−n=−nx−n−1\frac{d}{dx}x^{-n} = -nx^{-n-1}dxd​x−n=−nx−n−1

      Set u=1u = 1u=1 to get (1v)′=−v′v2\left(\frac{1}{v}\right)' = -\frac{v'}{v^2}(v1​)′=−v2v′​, then carry v=xnv = x^nv=xn through the exponent arithmetic to −nx−n−1-nx^{-n-1}−nx−n−1, and record that the power rule now covers negative nnn.

    Attempt 1 of 2