Lecture 9: Application of Derivatives to Kinematics, Product Rule, and Quotient Rule Application of Derivatives to Kinematics of Particles (3.3.3) Context: Lecture (9) for MATH 142-D01, conducted by Dr. Frank Nani on 3/11/20.
Core Concept: Using derivatives to analyze the movement of particles.
Variable Definitions:
Let s ( t ) s(t) s ( t ) t t t The particle moves in a rectilinear kinematic trajectory. Kinematic Equations:
(i) Instantaneous Velocity: The velocity v ( t ) v(t) v ( t ) t t t v ( t ) = d s d t v(t) = \frac{ds}{dt} v ( t ) = d t d s Speed: Speed is defined as the absolute value of velocity:
Speed = ∣ v ( t ) ∣ \text{Speed} = |v(t)| Speed = ∣ v ( t ) ∣ (ii) Average Velocity: In the time interval [ t 0 , t 2 ] [t_0, t_2] [ t 0 , t 2 ] v a v e v_{ave} v a v e v a v e = s ( t 2 ) − s ( t 1 ) t 2 − t 1 v_{ave} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} v a v e = t 2 − t 1 s ( t 2 ) − s ( t 1 ) (iii) Instantaneous Acceleration: The acceleration a ( t ) a(t) a ( t ) a ( t ) = d v d t a(t) = \frac{dv}{dt} a ( t ) = d t d v a ( t ) = d 2 s d t 2 a(t) = \frac{d^2s}{dt^2} a ( t ) = d t 2 d 2 s Worked Example: Kinematics
Equation of Motion: A particle is such that its position at any time is given by:
s ( t ) = t 4 − 2 t 3 + t 2 − t s(t) = t^4 - 2t^3 + t^2 - t s ( t ) = t 4 − 2 t 3 + t 2 − t (i) Find the velocity function: v ( t ) = d d t ( t 4 − 2 t 3 + t 2 − t ) v(t) = \frac{d}{dt}(t^4 - 2t^3 + t^2 - t) v ( t ) = d t d ( t 4 − 2 t 3 + t 2 − t ) v ( t ) = 4 t 3 − 6 t 2 + 2 t − 1 v(t) = 4t^3 - 6t^2 + 2t - 1 v ( t ) = 4 t 3 − 6 t 2 + 2 t − 1 (ii) Find the acceleration function: a ( t ) = d v d t a(t) = \frac{dv}{dt} a ( t ) = d t d v a ( t ) = d d t ( 4 t 3 − 6 t 2 + 2 t − 1 ) a(t) = \frac{d}{dt}(4t^3 - 6t^2 + 2t - 1) a ( t ) = d t d ( 4 t 3 − 6 t 2 + 2 t − 1 ) a ( t ) = 12 t 2 − 12 t + 2 a(t) = 12t^2 - 12t + 2 a ( t ) = 12 t 2 − 12 t + 2 (iii) Find the velocity at t = 1 t = 1 t = 1 t = 2 t = 2 t = 2 For t = 1 t = 1 t = 1 v ( 1 ) = 4 ( 1 ) 3 − 6 ( 1 ) 2 + 2 ( 1 ) − 1 v(1) = 4(1)^3 - 6(1)^2 + 2(1) - 1 v ( 1 ) = 4 ( 1 ) 3 − 6 ( 1 ) 2 + 2 ( 1 ) − 1 v ( 1 ) = 4 − 6 + 2 − 1 = − 1 m/s v(1) = 4 - 6 + 2 - 1 = -1\,\text{m/s} v ( 1 ) = 4 − 6 + 2 − 1 = − 1 m/s Note: The negative sign means the particle is moving to the left direction. For t = 2 t = 2 t = 2 v ( 2 ) = 4 ( 2 ) 3 − 6 ( 2 ) 2 + 2 ( 2 ) − 1 v(2) = 4(2)^3 - 6(2)^2 + 2(2) - 1 v ( 2 ) = 4 ( 2 ) 3 − 6 ( 2 ) 2 + 2 ( 2 ) − 1 v ( 2 ) = ( 4 ) ( 8 ) − ( 6 ) ( 4 ) + 4 − 1 v(2) = (4)(8) - (6)(4) + 4 - 1 v ( 2 ) = ( 4 ) ( 8 ) − ( 6 ) ( 4 ) + 4 − 1 v ( 2 ) = 32 − 24 + 4 − 1 = 11 m/s v(2) = 32 - 24 + 4 - 1 = 11\,\text{m/s} v ( 2 ) = 32 − 24 + 4 − 1 = 11 m/s Note: The positive sign means the particle is moving to the right direction. (iv) Find the acceleration at t = 1 t = 1 t = 1 a ( 1 ) = 12 ( 1 ) 2 − 12 ( 1 ) + 2 a(1) = 12(1)^2 - 12(1) + 2 a ( 1 ) = 12 ( 1 ) 2 − 12 ( 1 ) + 2 a ( 1 ) = 12 − 12 + 2 = 2 m/s 2 a(1) = 12 - 12 + 2 = 2\,\text{m/s}^2 a ( 1 ) = 12 − 12 + 2 = 2 m/s 2 The Product Rule (3.3.4) Definition: Let f ( x ) f(x) f ( x ) g ( x ) g(x) g ( x )
Formula: d d x [ f ( x ) ⋅ g ( x ) ] = f ( x ) ⋅ d d x [ g ( x ) ] + g ( x ) ⋅ d d x [ f ( x ) ] \frac{d}{dx}[f(x) \cdot g(x)] = f(x) \cdot \frac{d}{dx}[g(x)] + g(x) \cdot \frac{d}{dx}[f(x)] d x d [ f ( x ) ⋅ g ( x )] = f ( x ) ⋅ d x d [ g ( x )] + g ( x ) ⋅ d x d [ f ( x )]
Alternative Notations:
d d x [ f ( x ) ⋅ g ( x ) ] = f ′ ( x ) ⋅ g ( x ) + f ( x ) ⋅ g ′ ( x ) \frac{d}{dx}[f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x) d x d [ f ( x ) ⋅ g ( x )] = f ′ ( x ) ⋅ g ( x ) + f ( x ) ⋅ g ′ ( x ) d d x [ f ( x ) ⋅ g ( x ) ] = f ( x ) ⋅ g ′ ( x ) + g ( x ) ⋅ f ′ ( x ) \frac{d}{dx}[f(x) \cdot g(x)] = f(x) \cdot g'(x) + g(x) \cdot f'(x) d x d [ f ( x ) ⋅ g ( x )] = f ( x ) ⋅ g ′ ( x ) + g ( x ) ⋅ f ′ ( x ) Basic Principles: Differentiate one function, then multiply by the undifferentiated other function; then add the derivative of the other function multiplied by the first undifferentiated function.
Worked Example 1: Basic Product Rule
Function: f ( x ) = x e x f(x) = xe^x f ( x ) = x e x Strategy: This cannot be simplified by multiplication, so the product rule is applicable.Step #1: f ′ ( x ) = d d x ( x ⋅ e x ) f'(x) = \frac{d}{dx}(x \cdot e^x) f ′ ( x ) = d x d ( x ⋅ e x ) Step #2: Apply Rule: x ⋅ d d x ( e x ) + e x ⋅ d d x ( x ) x \cdot \frac{d}{dx}(e^x) + e^x \cdot \frac{d}{dx}(x) x ⋅ d x d ( e x ) + e x ⋅ d x d ( x ) Step #3: Evaluate derivatives: x ⋅ e x + e x ⋅ 1 x \cdot e^x + e^x \cdot 1 x ⋅ e x + e x ⋅ 1 Result: e x ( x + 1 ) e^x(x + 1) e x ( x + 1 ) Worked Example 2: Trigonometric and Exponential Functions
Function: F ( x ) = e 3 x ⋅ sin ( 4 x ) F(x) = e^{3x} \cdot \sin(4x) F ( x ) = e 3 x ⋅ sin ( 4 x ) Step #0: Identify f ( x ) = e 3 x f(x) = e^{3x} f ( x ) = e 3 x g ( x ) = sin ( 4 x ) g(x) = \sin(4x) g ( x ) = sin ( 4 x ) Step #1: F ′ ( x ) = e 3 x ⋅ d d x ( sin ( 4 x ) ) + sin ( 4 x ) ⋅ d d x ( e 3 x ) F'(x) = e^{3x} \cdot \frac{d}{dx}(\sin(4x)) + \sin(4x) \cdot \frac{d}{dx}(e^{3x}) F ′ ( x ) = e 3 x ⋅ d x d ( sin ( 4 x )) + sin ( 4 x ) ⋅ d x d ( e 3 x ) Step #2: Evaluate Chain Rule parts:d d x ( sin ( 4 x ) ) = cos ( 4 x ) ⋅ d d x ( 4 x ) = 4 cos ( 4 x ) \frac{d}{dx}(\sin(4x)) = \cos(4x) \cdot \frac{d}{dx}(4x) = 4\cos(4x) d x d ( sin ( 4 x )) = cos ( 4 x ) ⋅ d x d ( 4 x ) = 4 cos ( 4 x ) d d x ( e 3 x ) = e 3 x ⋅ d d x ( 3 x ) = 3 e 3 x \frac{d}{dx}(e^{3x}) = e^{3x} \cdot \frac{d}{dx}(3x) = 3e^{3x} d x d ( e 3 x ) = e 3 x ⋅ d x d ( 3 x ) = 3 e 3 x Step #3: Substitute back:
F ′ ( x ) = e 3 x ( 4 cos ( 4 x ) ) + sin ( 4 x ) ( 3 e 3 x ) F'(x) = e^{3x}(4\cos(4x)) + \sin(4x)(3e^{3x}) F ′ ( x ) = e 3 x ( 4 cos ( 4 x )) + sin ( 4 x ) ( 3 e 3 x ) F ′ ( x ) = 4 e 3 x cos ( 4 x ) + 3 e 3 x sin ( 4 x ) F'(x) = 4e^{3x}\cos(4x) + 3e^{3x}\sin(4x) F ′ ( x ) = 4 e 3 x cos ( 4 x ) + 3 e 3 x sin ( 4 x ) The Quotient Rule (3.3.5) Definition: Let f f f g g g
Formula: d d x [ f ( x ) g ( x ) ] = g ( x ) ⋅ d d x [ f ( x ) ] − f ( x ) ⋅ d d x [ g ( x ) ] [ g ( x ) ] 2 \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x) \cdot \frac{d}{dx}[f(x)] - f(x) \cdot \frac{d}{dx}[g(x)]}{[g(x)]^2} d x d [ g ( x ) f ( x ) ] = [ g ( x ) ] 2 g ( x ) ⋅ d x d [ f ( x )] − f ( x ) ⋅ d x d [ g ( x )]
Alternative Notation: g ( x ) ⋅ f ′ ( x ) − f ( x ) ⋅ g ′ ( x ) [ g ( x ) ] 2 \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{[g(x)]^2} [ g ( x ) ] 2 g ( x ) ⋅ f ′ ( x ) − f ( x ) ⋅ g ′ ( x )
Worked Example 1: Exponential and Power Function
Function: F ( x ) = e x 2 x 2 F(x) = \frac{e^x}{2x^2} F ( x ) = 2 x 2 e x Step #1: Factor out constant: F ′ ( x ) = 1 2 ⋅ d d x ( e x x 2 ) F'(x) = \frac{1}{2} \cdot \frac{d}{dx}\left(\frac{e^x}{x^2}\right) F ′ ( x ) = 2 1 ⋅ d x d ( x 2 e x ) Step #2: Apply Quotient Rule:
F ′ ( x ) = 1 2 ⋅ [ x 2 ⋅ d d x ( e x ) − e x ⋅ d d x ( x 2 ) ( x 2 ) 2 ] F'(x) = \frac{1}{2} \cdot \left[\frac{x^2 \cdot \frac{d}{dx}(e^x) - e^x \cdot \frac{d}{dx}(x^2)}{(x^2)^2}\right] F ′ ( x ) = 2 1 ⋅ [ ( x 2 ) 2 x 2 ⋅ d x d ( e x ) − e x ⋅ d x d ( x 2 ) ] F ′ ( x ) = 1 2 ⋅ [ x 2 e x − e x ( 2 x ) x 4 ] F'(x) = \frac{1}{2} \cdot \left[\frac{x^2e^x - e^x(2x)}{x^4}\right] F ′ ( x ) = 2 1 ⋅ [ x 4 x 2 e x − e x ( 2 x ) ] Step #3: Simplify (Numerator GCF is x x x F ′ ( x ) = 1 2 ⋅ [ x e x ( x − 2 ) x 4 ] F'(x) = \frac{1}{2} \cdot \left[\frac{xe^x(x - 2)}{x^4}\right] F ′ ( x ) = 2 1 ⋅ [ x 4 x e x ( x − 2 ) ] F ′ ( x ) = e x ( x − 2 ) 2 x 3 F'(x) = \frac{e^x(x - 2)}{2x^3} F ′ ( x ) = 2 x 3 e x ( x − 2 ) Worked Example 2: Rational Function
Function: f ( x ) = 3 x − 1 2 x + 1 f(x) = \frac{3x - 1}{2x + 1} f ( x ) = 2 x + 1 3 x − 1 Step #1: Identify parts: g ( x ) = 3 x − 1 → g ′ ( x ) = 3 g(x) = 3x - 1 \rightarrow g'(x) = 3 g ( x ) = 3 x − 1 → g ′ ( x ) = 3 h ( x ) = 2 x + 1 → h ′ ( x ) = 2 h(x) = 2x + 1 \rightarrow h'(x) = 2 h ( x ) = 2 x + 1 → h ′ ( x ) = 2 Step #2: Apply Rule:
f ′ ( x ) = ( 2 x + 1 ) ( 3 ) − ( 3 x − 1 ) ( 2 ) ( 2 x + 1 ) 2 f'(x) = \frac{(2x + 1)(3) - (3x - 1)(2)}{(2x + 1)^2} f ′ ( x ) = ( 2 x + 1 ) 2 ( 2 x + 1 ) ( 3 ) − ( 3 x − 1 ) ( 2 ) f ′ ( x ) = 6 x + 3 − ( 6 x − 2 ) ( 2 x + 1 ) 2 f'(x) = \frac{6x + 3 - (6x - 2)}{(2x + 1)^2} f ′ ( x ) = ( 2 x + 1 ) 2 6 x + 3 − ( 6 x − 2 ) f ′ ( x ) = 6 x + 3 − 6 x + 2 ( 2 x + 1 ) 2 f'(x) = \frac{6x + 3 - 6x + 2}{(2x + 1)^2} f ′ ( x ) = ( 2 x + 1 ) 2 6 x + 3 − 6 x + 2 Result: f ′ ( x ) = 5 ( 2 x + 1 ) 2 f'(x) = \frac{5}{(2x + 1)^2} f ′ ( x ) = ( 2 x + 1 ) 2 5 Worked Example 3: Trigonometric Quotient
Function: y = 3 x 8 − tan ( x ) y = \frac{3x}{8 - \tan(x)} y = 8 − t a n ( x ) 3 x Step #1: Identify f ( x ) = 3 x → f ′ ( x ) = 3 f(x) = 3x \rightarrow f'(x) = 3 f ( x ) = 3 x → f ′ ( x ) = 3 g ( x ) = 8 − tan ( x ) → g ′ ( x ) = − sec 2 ( x ) g(x) = 8 - \tan(x) \rightarrow g'(x) = -\sec^2(x) g ( x ) = 8 − tan ( x ) → g ′ ( x ) = − sec 2 ( x ) Step #2: Apply Rule:
y ′ ( x ) = ( 8 − tan ( x ) ) ( 3 ) − ( 3 x ) ( − sec 2 ( x ) ) ( 8 − tan ( x ) ) 2 y'(x) = \frac{(8 - \tan(x))(3) - (3x)(-\sec^2(x))}{(8 - \tan(x))^2} y ′ ( x ) = ( 8 − t a n ( x ) ) 2 ( 8 − t a n ( x )) ( 3 ) − ( 3 x ) ( − s e c 2 ( x )) y ′ ( x ) = 24 − 3 tan ( x ) + 3 x sec 2 ( x ) ( 8 − tan ( x ) ) 2 y'(x) = \frac{24 - 3\tan(x) + 3x\sec^2(x)}{(8 - \tan(x))^2} y ′ ( x ) = ( 8 − t a n ( x ) ) 2 24 − 3 t a n ( x ) + 3 x s e c 2 ( x ) Worked Example 4: Simplifying using GCF
Function: y = t ( t − 2 ) 2 y = \frac{t}{(t - 2)^2} y = ( t − 2 ) 2 t Step #1: Apply Quotient Rule:
y ′ ( t ) = ( t − 2 ) 2 ⋅ d d t ( t ) − t ⋅ d d t ( ( t − 2 ) 2 ) [ ( t − 2 ) 2 ] 2 y'(t) = \frac{(t - 2)^2 \cdot \frac{d}{dt}(t) - t \cdot \frac{d}{dt}((t - 2)^2)}{[(t - 2)^2]^2} y ′ ( t ) = [( t − 2 ) 2 ] 2 ( t − 2 ) 2 ⋅ d t d ( t ) − t ⋅ d t d (( t − 2 ) 2 ) y ′ ( t ) = ( t − 2 ) 2 ( 1 ) − t ( 2 ( t − 2 ) 1 ) ( t − 2 ) 4 y'(t) = \frac{(t - 2)^2(1) - t(2(t - 2)^1)}{(t - 2)^4} y ′ ( t ) = ( t − 2 ) 4 ( t − 2 ) 2 ( 1 ) − t ( 2 ( t − 2 ) 1 ) Step #2: Simplify: Identify t − 2 t - 2 t − 2 y ′ ( t ) = ( t − 2 ) ⋅ [ ( t − 2 ) − 2 t ] ( t − 2 ) 4 y'(t) = \frac{(t - 2) \cdot [(t - 2) - 2t]}{(t - 2)^4} y ′ ( t ) = ( t − 2 ) 4 ( t − 2 ) ⋅ [( t − 2 ) − 2 t ] Step #3: Cancel out ( t − 2 ) (t - 2) ( t − 2 ) y ′ ( t ) = t − 2 − 2 t ( t − 2 ) 3 y'(t) = \frac{t - 2 - 2t}{(t - 2)^3} y ′ ( t ) = ( t − 2 ) 3 t − 2 − 2 t y ′ ( t ) = − t − 2 ( t − 2 ) 3 y'(t) = \frac{-t - 2}{(t - 2)^3} y ′ ( t ) = ( t − 2 ) 3 − t − 2 Worked Example 5: Quadratic Denominator
Function: f ( t ) = 3 t t 2 + 1 f(t) = \frac{3t}{t^2 + 1} f ( t ) = t 2 + 1 3 t Calculation: f ′ ( t ) = ( t 2 + 1 ) ⋅ d d t ( 3 t ) − 3 t ⋅ d d t ( t 2 + 1 ) ( t 2 + 1 ) 2 f'(t) = \frac{(t^2 + 1) \cdot \frac{d}{dt}(3t) - 3t \cdot \frac{d}{dt}(t^2 + 1)}{(t^2 + 1)^2} f ′ ( t ) = ( t 2 + 1 ) 2 ( t 2 + 1 ) ⋅ d t d ( 3 t ) − 3 t ⋅ d t d ( t 2 + 1 ) f ′ ( t ) = ( t 2 + 1 ) ( 3 ) − 3 t ( 2 t ) ( t 2 + 1 ) 2 f'(t) = \frac{(t^2 + 1)(3) - 3t(2t)}{(t^2 + 1)^2} f ′ ( t ) = ( t 2 + 1 ) 2 ( t 2 + 1 ) ( 3 ) − 3 t ( 2 t ) f ′ ( t ) = 3 + 3 t 2 − 6 t 2 ( t 2 + 1 ) 2 f'(t) = \frac{3 + 3t^2 - 6t^2}{(t^2 + 1)^2} f ′ ( t ) = ( t 2 + 1 ) 2 3 + 3 t 2 − 6 t 2 f ′ ( t ) = 3 − 3 t 2 ( t 2 + 1 ) 2 f'(t) = \frac{3 - 3t^2}{(t^2 + 1)^2} f ′ ( t ) = ( t 2 + 1 ) 2 3 − 3 t 2 f ′ ( t ) = 3 ( 1 − t 2 ) ( t 2 + 1 ) 2 f'(t) = \frac{3(1 - t^2)}{(t^2 + 1)^2} f ′ ( t ) = ( t 2 + 1 ) 2 3 ( 1 − t 2 ) More Worked Examples (3.3.7) Product Rule Comparison (Methodical Check):
Function: f ( x ) = ( 1 + 3 x 2 ) ( x − x 2 ) f(x) = (1 + 3x^2)(x - x^2) f ( x ) = ( 1 + 3 x 2 ) ( x − x 2 ) Method (i): Multiplication/Expansion first: f ( x ) = x − x 2 + 3 x 3 − 3 x 4 f(x) = x - x^2 + 3x^3 - 3x^4 f ( x ) = x − x 2 + 3 x 3 − 3 x 4 f ( x ) = − 3 x 4 + 3 x 3 − x 2 + x f(x) = -3x^4 + 3x^3 - x^2 + x f ( x ) = − 3 x 4 + 3 x 3 − x 2 + x f ′ ( x ) = − 3 ( 4 ) x 3 + 3 ( 3 ) x 2 − 2 x + 1 f'(x) = -3(4)x^3 + 3(3)x^2 - 2x + 1 f ′ ( x ) = − 3 ( 4 ) x 3 + 3 ( 3 ) x 2 − 2 x + 1 f ′ ( x ) = − 12 x 3 + 9 x 2 − 2 x + 1 f'(x) = -12x^3 + 9x^2 - 2x + 1 f ′ ( x ) = − 12 x 3 + 9 x 2 − 2 x + 1 Method (ii): Product Rule application: f ′ ( x ) = ( 1 + 3 x 2 ) ⋅ d d x ( x − x 2 ) + ( x − x 2 ) ⋅ d d x ( 1 + 3 x 2 ) f'(x) = (1 + 3x^2) \cdot \frac{d}{dx}(x - x^2) + (x - x^2) \cdot \frac{d}{dx}(1 + 3x^2) f ′ ( x ) = ( 1 + 3 x 2 ) ⋅ d x d ( x − x 2 ) + ( x − x 2 ) ⋅ d x d ( 1 + 3 x 2 ) f ′ ( x ) = ( 1 + 3 x 2 ) ( 1 − 2 x ) + ( x − x 2 ) ( 6 x ) f'(x) = (1 + 3x^2)(1 - 2x) + (x - x^2)(6x) f ′ ( x ) = ( 1 + 3 x 2 ) ( 1 − 2 x ) + ( x − x 2 ) ( 6 x ) f ′ ( x ) = ( 1 − 2 x + 3 x 2 − 6 x 3 ) + ( 6 x 2 − 6 x 3 ) f'(x) = (1 - 2x + 3x^2 - 6x^3) + (6x^2 - 6x^3) f ′ ( x ) = ( 1 − 2 x + 3 x 2 − 6 x 3 ) + ( 6 x 2 − 6 x 3 ) f ′ ( x ) = − 12 x 3 + 9 x 2 − 2 x + 1 f'(x) = -12x^3 + 9x^2 - 2x + 1 f ′ ( x ) = − 12 x 3 + 9 x 2 − 2 x + 1 Worked Example with Radical and Exponential:
Function: f ( x ) = 3 e x x f(x) = 3e^x\sqrt{x} f ( x ) = 3 e x x Solution Rewrite: f ( x ) = 3 e x ⋅ x 1 / 2 f(x) = 3e^x \cdot x^{1/2} f ( x ) = 3 e x ⋅ x 1/2 Calculation: f ′ ( x ) = 3 ⋅ d d x ( e x ⋅ x 1 / 2 ) f'(x) = 3 \cdot \frac{d}{dx}(e^x \cdot x^{1/2}) f ′ ( x ) = 3 ⋅ d x d ( e x ⋅ x 1/2 ) f ′ ( x ) = 3 ⋅ [ e x ⋅ d d x ( x 1 / 2 ) + x 1 / 2 ⋅ d d x ( e x ) ] f'(x) = 3 \cdot [e^x \cdot \frac{d}{dx}(x^{1/2}) + x^{1/2} \cdot \frac{d}{dx}(e^x)] f ′ ( x ) = 3 ⋅ [ e x ⋅ d x d ( x 1/2 ) + x 1/2 ⋅ d x d ( e x )] f ′ ( x ) = 3 [ e x ⋅ 1 2 x − 1 / 2 + x 1 / 2 e x ] f'(x) = 3[e^x \cdot \frac{1}{2}x^{-1/2} + x^{1/2}e^x] f ′ ( x ) = 3 [ e x ⋅ 2 1 x − 1/2 + x 1/2 e x ] f ′ ( x ) = 3 e x [ 1 2 x + x ] f'(x) = 3e^x[\frac{1}{2\sqrt{x}} + \sqrt{x}] f ′ ( x ) = 3 e x [ 2 x 1 + x ] f ′ ( x ) = 3 e x [ 1 + 2 x 2 x ] f'(x) = 3e^x[\frac{1 + 2x}{2\sqrt{x}}] f ′ ( x ) = 3 e x [ 2 x 1 + 2 x ] f ′ ( x ) = 3 e x ( 1 + 2 x ) 2 x f'(x) = \frac{3e^x(1 + 2x)}{2\sqrt{x}} f ′ ( x ) = 2 x 3 e x ( 1 + 2 x ) Worked Example with Mixed Variables:
Function: R ( t ) = ( t + e t ) ( 3 − t ) R(t) = (t + e^t)(3 - \sqrt{t}) R ( t ) = ( t + e t ) ( 3 − t ) Calculation using Product Rule: R ′ ( t ) = ( t + e t ) ⋅ d d t ( 3 − t 1 / 2 ) + ( 3 − t 1 / 2 ) ⋅ d d t ( t + e t ) R'(t) = (t + e^t) \cdot \frac{d}{dt}(3 - t^{1/2}) + (3 - t^{1/2}) \cdot \frac{d}{dt}(t + e^t) R ′ ( t ) = ( t + e t ) ⋅ d t d ( 3 − t 1/2 ) + ( 3 − t 1/2 ) ⋅ d t d ( t + e t ) R ′ ( t ) = ( t + e t ) ( 0 − 1 2 t − 1 / 2 ) + ( 3 − t 1 / 2 ) ( 1 + e t ) R'(t) = (t + e^t)(0 - \frac{1}{2}t^{-1/2}) + (3 - t^{1/2})(1 + e^t) R ′ ( t ) = ( t + e t ) ( 0 − 2 1 t − 1/2 ) + ( 3 − t 1/2 ) ( 1 + e t ) R ′ ( t ) = − t 2 t − e t 2 t + 3 + 3 e t − t − t e t R'(t) = -\frac{t}{2\sqrt{t}} - \frac{e^t}{2\sqrt{t}} + 3 + 3e^t - \sqrt{t} - \sqrt{t}e^t R ′ ( t ) = − 2 t t − 2 t e t + 3 + 3 e t − t − t e t R ′ ( t ) = − 1 2 t − e t 2 t + 3 + 3 e t − t − t e t R'(t) = -\frac{1}{2}\sqrt{t} - \frac{e^t}{2\sqrt{t}} + 3 + 3e^t - \sqrt{t} - \sqrt{t}e^t R ′ ( t ) = − 2 1 t − 2 t e t + 3 + 3 e t − t − t e t R ′ ( t ) = 3 − 3 2 t + ( 3 − 1 2 t − t ) e t R'(t) = 3 - \frac{3}{2}\sqrt{t} + (3 - \frac{1}{2\sqrt{t}} - \sqrt{t})e^t R ′ ( t ) = 3 − 2 3 t + ( 3 − 2 t 1 − t ) e t