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Alex Ar [27]
3 years ago
8

Find

y}{dx} " align="absmiddle" class="latex-formula">
if
y = (x +  \sqrt{x} )^{2}
​
Mathematics
1 answer:
nataly862011 [7]3 years ago
4 0

Answer:

\displaystyle y' = 2x + 3\sqrt{x} + 1

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>

  • Terms/Coefficients
  • Anything to the 0th power is 1
  • Exponential Rule [Rewrite]:                                                                              \displaystyle b^{-m} = \frac{1}{b^m}
  • Exponential Rule [Root Rewrite]:                                                                     \displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}<u> </u>

<u>Calculus</u>

Derivatives

Derivative Notation

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)

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

<em />\displaystyle y = (x + \sqrt{x})^2<em />

<em />

<u>Step 2: Differentiate</u>

  1. Chain Rule:                                                                                                        \displaystyle y' = 2(x + \sqrt{x})^{2 - 1} \cdot \frac{d}{dx}[x + \sqrt{x}]
  2. Rewrite [Exponential Rule - Root Rewrite]:                                                     \displaystyle y' = 2(x + x^{\frac{1}{2}})^{2 - 1} \cdot \frac{d}{dx}[x + x^{\frac{1}{2}}]
  3. Simplify:                                                                                                             \displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot \frac{d}{dx}[x + x^{\frac{1}{2}}]
  4. Basic Power Rule:                                                                                             \displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot (1 \cdot x^{1 - 1} + \frac{1}{2}x^{\frac{1}{2} - 1})
  5. Simplify:                                                                                                             \displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot (1 + \frac{1}{2}x^{-\frac{1}{2}})
  6. Rewrite [Exponential Rule - Rewrite]:                                                              \displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot (1 + \frac{1}{2x^{\frac{1}{2}}})
  7. Multiply:                                                                                                             \displaystyle y' = 2[(x + x^{\frac{1}{2}}) + \frac{x + x^{\frac{1}{2}}}{2x^{\frac{1}{2}}}]
  8. [Brackets] Add:                                                                                                 \displaystyle y' = 2(\frac{2x + 3x^{\frac{1}{2}} + 1}{2})
  9. Multiply:                                                                                                             \displaystyle y' = 2x + 3x^{\frac{1}{2}} + 1
  10. Rewrite [Exponential Rule - Root Rewrite]:                                                     \displaystyle y' = 2x + 3\sqrt{x} + 1

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

Unit: Derivatives

Book: College Calculus 10e

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