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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∞

The Power Rule for Rational Exponents

How does turning y = x^(m/n) into y^n = x^m let the chain rule prove the power rule for every rational exponent?


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    1. 1

      Setting up yn=xmy^n = x^myn=xm and differentiating it

      Write y=xm/ny = x^{m/n}y=xm/n, raise it to the nnnth power to get yn=xmy^n = x^myn=xm, note which equation gets differentiated, and apply the chain rule to reach nyn−1dydx=mxm−1ny^{n-1}\frac{dy}{dx} = mx^{m-1}nyn−1dxdy​=mxm−1.

    2. 2

      Solving for dydx\frac{dy}{dx}dxdy​ and the exponent arithmetic

      Divide to isolate dydx\frac{dy}{dx}dxdy​, substitute y=xm/ny = x^{m/n}y=xm/n back in, and carry out the exponent algebra (m−1)−m(n−1)n=mn−1(m-1) - \frac{m(n-1)}{n} = \frac{m}{n} - 1(m−1)−nm(n−1)​=nm​−1 to land on axa−1ax^{a-1}axa−1.

    3. 3

      The case m=1m = 1m=1 and the nnnth roots

      Record that a=mna = \frac{m}{n}a=nm​ with m=1m = 1m=1 gives x1/nx^{1/n}x1/n, and differentiate one root such as x\sqrt{x}x​ or x1/3x^{1/3}x1/3 as the worked example.

    Attempt 1 of 2