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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 Chain Rule: Differentiating a Function of a Function

Why is the derivative of a composition a product, and what has to happen to the inside variable for the cancellation to survive the limit?


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Your summary note

    1. 1

      Composition as substitution, with a named intermediate variable

      Write (sin⁡t)10(\sin t)^{10}(sint)10 as two steps, x=sin⁡tx = \sin tx=sint then y=x10y = x^{10}y=x10, note the sin⁡10t\sin^{10} tsin10t shorthand, and record that the last step substitutes xxx back.

    2. 2

      The one-line derivation of dy/dt=(dy/dx)(dx/dt)dy/dt = (dy/dx)(dx/dt)dy/dt=(dy/dx)(dx/dt)

      Write Δy/Δt=(Δy/Δx)(Δx/Δt)\Delta y/\Delta t = (\Delta y/\Delta x)(\Delta x/\Delta t)Δy/Δt=(Δy/Δx)(Δx/Δt), cancel Δx\Delta xΔx, record the limit step as Δt→0\Delta t\to0Δt→0, and state both dy/dt=(dy/dx)(dx/dt)dy/dt = (dy/dx)(dx/dt)dy/dt=(dy/dx)(dx/dt) and ddtf(g(t))=f′(g(t))g′(t)\frac{d}{dt}f(g(t)) = f'(g(t))g'(t)dtd​f(g(t))=f′(g(t))g′(t).

    3. 3

      Worked example (sin⁡t)10(\sin t)^{10}(sint)10

      Take dy/dx=10x9dy/dx = 10x^9dy/dx=10x9 and dx/dt=cos⁡tdx/dt = \cos tdx/dt=cost, multiply them, and substitute x=sin⁡tx = \sin tx=sint back to finish at 10sin⁡9tcos⁡t10\sin^9 t\cos t10sin9tcost.

    4. 4

      sin⁡(10t)\sin(10t)sin(10t) and the shortcut without the middle variable

      With x=10tx = 10tx=10t and y=sin⁡xy = \sin xy=sinx, reach 10cos⁡(10t)10\cos(10t)10cos(10t), then redo it without naming xxx, differentiating the outside function, leaving the inside expression alone, and multiplying by its derivative.

    Attempt 1 of 2