Day 22 - The Chain Rule + Applied Differentiation Notes Course Overview and Learning Objectives Foundations and Mathematical Formulation of the Chain Rule Theoretical Statement: If f f f u u u u u u x x x f ( u ) f(u) f ( u ) x x x
Leibniz Notation:
Lagrange / Functional Notation:
Core Mechanism: Rate of change of composite function = (Derivative of outer evaluated at inner) × \times ×
Example 22.1: Logarithmic, Absolute Value, and Algebraic Derivatives Part (a): Differentiate f ( x ) = log 3 ( x 2 ) f(x) = \log_3(x^2) f ( x ) = log 3 ( x 2 )
Outer: g ( u ) = log 3 ( u ) ⟹ d g d u = 1 u ln ( 3 ) g(u) = \log_3(u) \implies \frac{dg}{du} = \frac{1}{u \ln(3)} g ( u ) = log 3 ( u ) ⟹ d u d g = u l n ( 3 ) 1
Inner: u = x 2 ⟹ d u d x = 2 x u = x^2 \implies \frac{du}{dx} = 2x u = x 2 ⟹ d x d u = 2 x
Chain Rule: f ′ ( x ) = 1 x 2 ln ( 3 ) ⋅ 2 x = 2 x ln ( 3 ) f'(x) = \frac{1}{x^2 \ln(3)} \cdot 2x = \frac{2}{x \ln(3)} f ′ ( x ) = x 2 l n ( 3 ) 1 ⋅ 2 x = x l n ( 3 ) 2 x ≠ 0 x \neq 0 x = 0
Log Rule Alternative: f ( x ) = 2 log 3 ( ∣ x ∣ ) ⟹ f ′ ( x ) = 2 ⋅ 1 x ln ( 3 ) = 2 x ln ( 3 ) f(x) = 2\log_3(|x|) \implies f'(x) = 2 \cdot \frac{1}{x \ln(3)} = \frac{2}{x \ln(3)} f ( x ) = 2 log 3 ( ∣ x ∣ ) ⟹ f ′ ( x ) = 2 ⋅ x l n ( 3 ) 1 = x l n ( 3 ) 2
Part (b): Differentiate f ( x ) = log 3 ( x 2 + 1 ) f(x) = \log_3(x^2 + 1) f ( x ) = log 3 ( x 2 + 1 )
Outer: g ( u ) = log 3 ( u ) ⟹ d g d u = 1 u ln ( 3 ) g(u) = \log_3(u) \implies \frac{dg}{du} = \frac{1}{u \ln(3)} g ( u ) = log 3 ( u ) ⟹ d u d g = u l n ( 3 ) 1
Inner: u = x 2 + 1 ⟹ d u d x = 2 x u = x^2 + 1 \implies \frac{du}{dx} = 2x u = x 2 + 1 ⟹ d x d u = 2 x
Result: f ′ ( x ) = 1 ( x 2 + 1 ) ln ( 3 ) ⋅ 2 x = 2 x ( x 2 + 1 ) ln ( 3 ) f'(x) = \frac{1}{(x^2 + 1)\ln(3)} \cdot 2x = \frac{2x}{(x^2 + 1)\ln(3)} f ′ ( x ) = ( x 2 + 1 ) l n ( 3 ) 1 ⋅ 2 x = ( x 2 + 1 ) l n ( 3 ) 2 x
Domain Note: Because x 2 + 1 ≥ 1 > 0 x^2 + 1 \ge 1 > 0 x 2 + 1 ≥ 1 > 0 x ∈ R x \in \mathbb{R} x ∈ R ( − ∞ , ∞ ) (-\infty, \infty) ( − ∞ , ∞ )
Part (c): Differentiate the Absolute Value Function y = ∣ x ∣ y = |x| y = ∣ x ∣
Radical Form: y = x 2 = ( x 2 ) 1 / 2 y = \sqrt{x^2} = (x^2)^{1/2} y = x 2 = ( x 2 ) 1/2
Outer: f ( u ) = u 1 / 2 ⟹ d f d u = 1 2 u f(u) = u^{1/2} \implies \frac{df}{du} = \frac{1}{2\sqrt{u}} f ( u ) = u 1/2 ⟹ d u df = 2 u 1
Inner: u = x 2 ⟹ d u d x = 2 x u = x^2 \implies \frac{du}{dx} = 2x u = x 2 ⟹ d x d u = 2 x
Assembly: d y d x = 1 2 x 2 ⋅ 2 x = x x 2 = x ∣ x ∣ = ∣ x ∣ x = sgn ( x ) \frac{dy}{dx} = \frac{1}{2\sqrt{x^2}} \cdot 2x = \frac{x}{\sqrt{x^2}} = \frac{x}{|x|} = \frac{|x|}{x} = \text{sgn}(x) d x d y = 2 x 2 1 ⋅ 2 x = x 2 x = ∣ x ∣ x = x ∣ x ∣ = sgn ( x )
Piecewise Evaluation:
For x > 0 x > 0 x > 0 d y d x = 1 \frac{dy}{dx} = 1 d x d y = 1
For x < 0 x < 0 x < 0 d y d x = − 1 \frac{dy}{dx} = -1 d x d y = − 1
At x = 0 x = 0 x = 0 undefined (corner/cusp; left-hand derivative − 1 ≠ -1 \neq − 1 = 1 1 1
Part (d): Differentiate h ( x ) = 1 3 x 2 + 1 h(x) = \frac{1}{\sqrt{3x^2 + 1}} h ( x ) = 3 x 2 + 1 1
Exponential Form: h ( x ) = ( 3 x 2 + 1 ) − 1 / 2 h(x) = (3x^2 + 1)^{-1/2} h ( x ) = ( 3 x 2 + 1 ) − 1/2
Outer: f ( u ) = u − 1 / 2 ⟹ d f d u = − 1 2 u − 3 / 2 f(u) = u^{-1/2} \implies \frac{df}{du} = -\frac{1}{2}u^{-3/2} f ( u ) = u − 1/2 ⟹ d u df = − 2 1 u − 3/2
Inner: u = 3 x 2 + 1 ⟹ d u d x = 6 x u = 3x^2 + 1 \implies \frac{du}{dx} = 6x u = 3 x 2 + 1 ⟹ d x d u = 6 x
Assembly: h ′ ( x ) = − 1 2 ( 3 x 2 + 1 ) − 3 / 2 ⋅ 6 x = − 3 x ( 3 x 2 + 1 ) 3 / 2 = − 3 x ( 3 x 2 + 1 ) 3 h'(x) = -\frac{1}{2}(3x^2 + 1)^{-3/2} \cdot 6x = -\frac{3x}{(3x^2 + 1)^{3/2}} = -\frac{3x}{\sqrt{(3x^2 + 1)^3}} h ′ ( x ) = − 2 1 ( 3 x 2 + 1 ) − 3/2 ⋅ 6 x = − ( 3 x 2 + 1 ) 3/2 3 x = − ( 3 x 2 + 1 ) 3 3 x
Part (e): Differentiate k ( t ) = ( 5 t 1 / 4 + 9 t + 7 ) 20 k(t) = (5t^{1/4} + 9t + 7)^{20} k ( t ) = ( 5 t 1/4 + 9 t + 7 ) 20
Generalized Power Rule: d d t [ u n ] = n u n − 1 d u d t \frac{d}{dt}[u^n] = n u^{n-1} \frac{du}{dt} d t d [ u n ] = n u n − 1 d t d u
Inner Function: u ( t ) = 5 t 1 / 4 + 9 t + 7 ⟹ d u d t = 5 4 t − 3 / 4 + 9 u(t) = 5t^{1/4} + 9t + 7 \implies \frac{du}{dt} = \frac{5}{4}t^{-3/4} + 9 u ( t ) = 5 t 1/4 + 9 t + 7 ⟹ d t d u = 4 5 t − 3/4 + 9
Chain Rule: k ′ ( t ) = 20 ( 5 t 1 / 4 + 9 t + 7 ) 19 ( 5 4 t − 3 / 4 + 9 ) = ( 5 t 1 / 4 + 9 t + 7 ) 19 ( 25 t − 3 / 4 + 180 ) k'(t) = 20(5t^{1/4} + 9t + 7)^{19} \left(\frac{5}{4}t^{-3/4} + 9\right) = (5t^{1/4} + 9t + 7)^{19}(25t^{-3/4} + 180) k ′ ( t ) = 20 ( 5 t 1/4 + 9 t + 7 ) 19 ( 4 5 t − 3/4 + 9 ) = ( 5 t 1/4 + 9 t + 7 ) 19 ( 25 t − 3/4 + 180 )
Example 22.2: Marginal Profit and Labor Production Application Example 22.3: Exponential, Logistic, and Variable-Base Functions Part (a): Exponential with Linear Power f ( x ) = e k x f(x) = e^{kx} f ( x ) = e k x
Part (b): Gaussian Density Function f ( x ) = 1 2 π e − x 2 / 2 f(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2} f ( x ) = 2 π 1 e − x 2 /2
Inner Exponent: u = − 1 2 x 2 ⟹ d u d x = − x u = -\frac{1}{2}x^2 \implies \frac{du}{dx} = -x u = − 2 1 x 2 ⟹ d x d u = − x
Chain Rule: f ′ ( x ) = 1 2 π e − x 2 / 2 ⋅ ( − x ) = − x 2 π e − x 2 / 2 f'(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \cdot (-x) = -\frac{x}{\sqrt{2\pi}} e^{-x^2/2} f ′ ( x ) = 2 π 1 e − x 2 /2 ⋅ ( − x ) = − 2 π x e − x 2 /2
Part (c): Logistic Function P ( t ) = N 1 + C b − t = N ( 1 + C b − t ) − 1 P(t) = \frac{N}{1 + Cb^{-t}} = N(1 + Cb^{-t})^{-1} P ( t ) = 1 + C b − t N = N ( 1 + C b − t ) − 1
Outer: g ( u ) = N u − 1 ⟹ d g d u = − N u 2 g(u) = N u^{-1} \implies \frac{dg}{du} = -\frac{N}{u^2} g ( u ) = N u − 1 ⟹ d u d g = − u 2 N
Inner: u = 1 + C b − t ⟹ d u d t = − C b − t ln ( b ) u = 1 + Cb^{-t} \implies \frac{du}{dt} = -C b^{-t} \ln(b) u = 1 + C b − t ⟹ d t d u = − C b − t ln ( b )
Chain Rule: P ′ ( t ) = ( − N ( 1 + C b − t ) 2 ) ⋅ ( − C b − t ln ( b ) ) = N C b − t ln ( b ) ( 1 + C b − t ) 2 P'(t) = \left(-\frac{N}{(1 + Cb^{-t})^2}\right) \cdot \left(-C b^{-t}\ln(b)\right) = \frac{N C b^{-t}\ln(b)}{(1 + Cb^{-t})^2} P ′ ( t ) = ( − ( 1 + C b − t ) 2 N ) ⋅ ( − C b − t ln ( b ) ) = ( 1 + C b − t ) 2 N C b − t l n ( b )
Part (d): Exponential Polynomial g ( t ) = e t 5 − 5 t 4 + 10 t 3 − 10 t 2 + 5 t − 1 g(t) = e^{t^5 - 5t^4 + 10t^3 - 10t^2 + 5t - 1} g ( t ) = e t 5 − 5 t 4 + 10 t 3 − 10 t 2 + 5 t − 1
Binomial Factorization: Exponent matches ( t − 1 ) 5 (t - 1)^5 ( t − 1 ) 5 g ( t ) = e ( t − 1 ) 5 g(t) = e^{(t - 1)^5} g ( t ) = e ( t − 1 ) 5
Derivative of Inner: u ′ ( t ) = 5 ( t − 1 ) 4 = 5 t 4 − 20 t 3 + 30 t 2 − 20 t + 5 u'(t) = 5(t - 1)^4 = 5t^4 - 20t^3 + 30t^2 - 20t + 5 u ′ ( t ) = 5 ( t − 1 ) 4 = 5 t 4 − 20 t 3 + 30 t 2 − 20 t + 5
Factored Form: g ′ ( t ) = 5 ( t − 1 ) 4 e ( t − 1 ) 5 g'(t) = 5(t - 1)^4 e^{(t - 1)^5} g ′ ( t ) = 5 ( t − 1 ) 4 e ( t − 1 ) 5
Expanded Form: g ′ ( t ) = ( 5 t 4 − 20 t 3 + 30 t 2 − 20 t + 5 ) e t 5 − 5 t 4 + 10 t 3 − 10 t 2 + 5 t − 1 g'(t) = (5t^4 - 20t^3 + 30t^2 - 20t + 5) e^{t^5 - 5t^4 + 10t^3 - 10t^2 + 5t - 1} g ′ ( t ) = ( 5 t 4 − 20 t 3 + 30 t 2 − 20 t + 5 ) e t 5 − 5 t 4 + 10 t 3 − 10 t 2 + 5 t − 1
Part (e): Variable-Base Variable-Exponent Function y = x x y = x^x y = x x x > 0 x > 0 x > 0
Method 1 (Base e e e
Rewrite: y = e ln ( x x ) = e x ln ( x ) y = e^{\ln(x^x)} = e^{x\ln(x)} y = e l n ( x x ) = e x l n ( x )
Inner Product Rule: d d x [ x ln ( x ) ] = 1 ⋅ ln ( x ) + x ⋅ 1 x = ln ( x ) + 1 \frac{d}{dx}[x\ln(x)] = 1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1 d x d [ x ln ( x )] = 1 ⋅ ln ( x ) + x ⋅ x 1 = ln ( x ) + 1
Chain Rule: d y d x = e x ln ( x ) ( ln ( x ) + 1 ) = x x ( ln ( x ) + 1 ) \frac{dy}{dx} = e^{x\ln(x)}(\ln(x) + 1) = x^x(\ln(x) + 1) d x d y = e x l n ( x ) ( ln ( x ) + 1 ) = x x ( ln ( x ) + 1 )
Method 2 (Implicit Logarithmic Differentiation):
Take ln \ln ln ln ( y ) = x ln ( x ) \ln(y) = x\ln(x) ln ( y ) = x ln ( x )
Implicit Derivative: 1 y d y d x = ln ( x ) + 1 ⟹ d y d x = y ( ln ( x ) + 1 ) = x x ( ln ( x ) + 1 ) \frac{1}{y}\frac{dy}{dx} = \ln(x) + 1 \implies \frac{dy}{dx} = y(\ln(x) + 1) = x^x(\ln(x) + 1) y 1 d x d y = ln ( x ) + 1 ⟹ d x d y = y ( ln ( x ) + 1 ) = x x ( ln ( x ) + 1 )
Summary and Academic Deliverables