Question

A function is given by f(x) =
$\sqrt[4]{x}$
Evaluate f'(16).​

Answer 1

Answer:

$\displaystyle f'(16) = \frac{1}{32}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Algebra I

• Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
• Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

Step 1: Define

$\displaystyle f(x) = \sqrt[4]{x}$

f'(16) is x = 16 for the derivative f'(x)

Step 2: Differentiate

1. [Function] Rewrite [Exponential Rule - Root Rewrite]:                               $\displaystyle f(x) = x^{\frac{1}{4}}$
2. Basic Power Rule:                                                                                         $\displaystyle f'(x) = \frac{1}{4}x^{\frac{1}{4} - 1}$
3. Simplify:                                                                                                         $\displaystyle f'(x) = \frac{1}{4}x^{\frac{-3}{4}}$
4. Rewrite [Exponential Rule - Rewrite]:                                                           $\displaystyle f'(x) = \frac{1}{4x^{\frac{3}{4}}}$

Step 3: Solve

1. Substitute in x [Derivative]:                                                                           $\displaystyle f'(16) = \frac{1}{4(16)^{\frac{3}{4}}}$
2. Evaluate exponents:                                                                                     $\displaystyle f'(16) = \frac{1}{4(8)}$
3. Multiply:                                                                                                         $\displaystyle f'(16) = \frac{1}{32}$

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

Unit: Derivatives

Book: College Calculus 10e