Mathematics, 02.07.2019 09:30 ivymckinney21

Use the chain rule to differentiate the following function. f(x)=(5x+sin^3(x)+sinx^3)^3

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Answer from: bridgette1227

$\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \bigg[ 3x^2cosx^3 + 3sin^2(x)cos(x) + 5 \bigg]$

General Formulas and Concepts:

Algebra I

FunctionsFunction Notation

Pre-Calculus

Trigonometric Notation

Calculus

Derivatives

Derivative Notation

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Trig Derivative: $\displaystyle \frac{d}{dx}[sin(u)] = u'cos(u)$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = (5x + sin^3x + sinx^3)^3$

Step 2: Differentiate

Chain Rule: $\displaystyle f'(x) = \frac{d}{dx} \bigg[ (5x + sin^3x + sinx^3)^3 \bigg] \cdot \frac{d}{dx} \bigg[ 5x + sin^3x + sinx^3 \bigg]$ Basic Power Rule: $\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^{3 - 1} \cdot \frac{d}{dx} \bigg[ 5x + sin^3x + sinx^3 \bigg]$ Simplify: $\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \frac{d}{dx} \bigg[ 5x + sin^3x + sinx^3 \bigg]$ Derivative Property [Addition]: $\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ \frac{d}{dx}[5x] + \frac{d}{dx}[sin^3x] + \frac{d}{dx}[sinx^3] \bigg]$ Rewrite [Trigonometric Notation]: $\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ \frac{d}{dx}[5x] + \frac{d}{dx}[(sinx)^3] + \frac{d}{dx}[sinx^3] \bigg]$ Basic Power Rule: $\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ 5x^{1 - 1} + \frac{d}{dx}[(sinx)^3] + \frac{d}{dx}[sinx^3] \bigg]$ Simplify: $\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ 5 + \frac{d}{dx}[(sinx)^3] + \frac{d}{dx}[sinx^3] \bigg]$ Trig Derivative [Derivative Rule - Chain Rule]: $\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ 5 + \frac{d}{dx}[(sinx)^3] + \bigg( \frac{d}{dx}[sinx^3] \cdot \frac{d}{dx}[x^3] \bigg) \bigg]$ Trig Derivative [Basic Power Rule]: $\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ 5 + \frac{d}{dx}[(sinx)^3] + \bigg( cosx^3 \cdot 3x^{3 - 1} \bigg) \bigg]$ Simplify: $\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ 5 + \frac{d}{dx}[(sinx)^3] + 3x^2cosx^3 \bigg]$ Chain Rule: $\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ 5 + \bigg( \frac{d}{dx}[(sinx)^3] \cdot \frac{d}{dx}[sinx] \bigg) + 3x^2cosx^3 \bigg]$ Basic Power Rule: $\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ 5 + \bigg( 3(sinx)^{3 - 1} \cdot \frac{d}{dx}[sinx] \bigg) + 3x^2cosx^3 \bigg]$ Simplify: $\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ 5 + \bigg( 3(sinx)^2 \cdot \frac{d}{dx}[sinx] \bigg) + 3x^2cosx^3 \bigg]$ Trig Derivative: $\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ 5 + \bigg( 3(sinx)^2 \cdot cos(x) \bigg) + 3x^2cosx^3 \bigg]$ Simplify [Rewrite]: $\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ 5 + 3sin^2(x)cos(x) + 3x^2cosx^3 \bigg]$ Rewrite: $\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \bigg[ 3x^2cosx^3 + 3sin^2(x)cos(x) + 5 \bigg]$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Derivatives

Book: College Calculus 10e

Answer from: Quest

The answer is 6 weeks

Answer from: Quest

the length of o'p' is $1.5\times op$ .