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ExtremeBDS [4]
3 years ago
11

Find the derivative of

\sqrt[3]{x} " alt=" y = \sqrt[3]{x} " align="absmiddle" class="latex-formula">
at the point where x = 27.​
Mathematics
1 answer:
MAXImum [283]3 years ago
8 0

Answer:

\displaystyle y'(27) = \frac{1}{27}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Algebra I</u>

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

<u>Calculus</u>

Derivatives

Derivative Notation

Basic Power Rule

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

<u>Step 1: Define</u>

y = ∛x

x = 27

<u>Step 2: Differentiate</u>

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

<u>Step 3: Solve</u>

  1. Substitute<em> </em>in <em>x</em> [Derivative]:                                                                           \displaystyle y'(27) = \frac{1}{3(27)^{\frac{2}{3}}}
  2. Evaluate exponents:                                                                                     \displaystyle y'(27) = \frac{1}{3(9)}
  3. Multiply:                                                                                                         \displaystyle y'(27) = \frac{1}{27}

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

Unit: Derivatives

Book: College Calculus 10e

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Answer:

<em>It will occur zero times between midnight and one o'clock.</em>

Step-by-step explanation:

<u>Least Common Multiple (LCM)</u>

Three events keep James from sleeping: his clock ticking every 20 seconds, a tap dripping every 15 seconds, and his dog snoring every 27 seconds.

All three events happened together at midnight. They will happen together again the first time the numbers 20, 15, and 27 have a common multiple. This is the LCM.

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Now multiply all the factors the maximum number of times they appear:

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3 years ago
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Answer:

Not factorable.

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4t= 1t+-7t doesn't work

4t=-1t+7t doesn't work

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7 0
3 years ago
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