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ExtremeBDS [4]
2 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]2 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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-1/3

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3 years ago
What type of triangle has side lengths 9, 10, and √130? A. obtuse B. not a triangle C. acute D. right
EastWind [94]

Answer: Option C.

Step-by-step explanation:

The lengths of our triangle are:

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If the triangle is a triangle rectangle, by the Pitagoream's theorem we have:

A^2 + B^2 = H^2

in this case H is the larger side, this must be √130.

then:

A and B must be 9 and 10.

9^2 + 10^2 = (√130)^2

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This is false, so this is NOT a triangle rectangle, the hypotenuse is shorter than it should be.

Now, we have some kind of rule:

 if A^2 + B^2 = H^2 then we have one angle of 90° and two smaller ones. (triangle rectangle)

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2 years ago
Evauluate 37/100+3/10
zheka24 [161]

Answer:

67/100

Step-by-step explanation:

Find common denominators, note that what you do to the denominator, you must do to the numerator.

The common denominator is 100:

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67/100 is your answer.

~

4 0
3 years ago
Read 2 more answers
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