Question

Use the chain rule to differentiate the following function.

f(x)=(5x+sin^3(x)+sinx^3)^3

Answer 1

`\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

• Functions
• Function Notation

Pre-Calculus

• Trigonometric Notation

Calculus

Derivatives

Derivative Notation

Derivative Property [Addition/Subtraction]:
`\displaystyle (d)/(dx)[f(x) + g(x)] = (d)/(dx)[f(x)] + (d)/(dx)[g(x)]`

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
`\displaystyle (d)/(dx)[f(g(x))] =f'(g(x)) \cdot g'(x)`

Trig Derivative:
`\displaystyle (d)/(dx)[sin(u)] = u'cos(u)`

Explanation:

Step 1: Define

Identify

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

Step 2: Differentiate

1. Chain Rule:
   `\displaystyle f'(x) = (d)/(dx) \bigg[ (5x + sin^3x + sinx^3)^3 \bigg] \cdot (d)/(dx) \bigg[ 5x + sin^3x + sinx^3 \bigg]`
2. Basic Power Rule:
   `\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^(3 - 1) \cdot (d)/(dx) \bigg[ 5x + sin^3x + sinx^3 \bigg]`
3. Simplify:
   `\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot (d)/(dx) \bigg[ 5x + sin^3x + sinx^3 \bigg]`
4. Derivative Property [Addition]:
   `\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ (d)/(dx)[5x] + (d)/(dx)[sin^3x] + (d)/(dx)[sinx^3] \bigg]`
5. Rewrite [Trigonometric Notation]:
   `\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ (d)/(dx)[5x] + (d)/(dx)[(sinx)^3] + (d)/(dx)[sinx^3] \bigg]`
6. Basic Power Rule:
   `\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ 5x^(1 - 1) + (d)/(dx)[(sinx)^3] + (d)/(dx)[sinx^3] \bigg]`
7. Simplify:
   `\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ 5 + (d)/(dx)[(sinx)^3] + (d)/(dx)[sinx^3] \bigg]`
8. Trig Derivative [Derivative Rule - Chain Rule]:
   `\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ 5 + (d)/(dx)[(sinx)^3] + \bigg( (d)/(dx)[sinx^3] \cdot (d)/(dx)[x^3] \bigg) \bigg]`
9. Trig Derivative [Basic Power Rule]:
   `\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ 5 + (d)/(dx)[(sinx)^3] + \bigg( cosx^3 \cdot 3x^(3 - 1) \bigg) \bigg]`
10. Simplify:
    `\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ 5 + (d)/(dx)[(sinx)^3] + 3x^2cosx^3 \bigg]`
11. Chain Rule:
    `\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ 5 + \bigg( (d)/(dx)[(sinx)^3] \cdot (d)/(dx)[sinx] \bigg) + 3x^2cosx^3 \bigg]`
12. Basic Power Rule:
    `\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ 5 + \bigg( 3(sinx)^(3 - 1) \cdot (d)/(dx)[sinx] \bigg) + 3x^2cosx^3 \bigg]`
13. Simplify:
    `\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ 5 + \bigg( 3(sinx)^2 \cdot (d)/(dx)[sinx] \bigg) + 3x^2cosx^3 \bigg]`
14. 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]`
15. Simplify [Rewrite]:
    `\displaystyle f'(x) = 3(5x + sin^3x + sinx^3)^2 \cdot \bigg[ 5 + 3sin^2(x)cos(x) + 3x^2cosx^3 \bigg]`
16. 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