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What is the slope of the line tangent to the curve square root (x) +square root (y) = 2 at the point ( 9/4, 1/4 )? (photo attached of answer choices)

What is the slope of the line tangent to the curve square root x square root y 2 at the point 94 14 photo attached of answer choices class=

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Answer:

B. [tex]\displaystyle -\frac{1}{3}[/tex]

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  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality

Algebra I

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

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

The definition of a derivative is the slope of the tangent line

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Implicit Differentiation

Step-by-step explanation:

Step 1: Define

[tex]\displaystyle \sqrt{x} + \sqrt{y} = 2[/tex]

[tex]\displaystyle (\frac{9}{4}, \frac{1}{4})[/tex]

Step 2: Differentiate

Implicit Differentiation

  1. [Function] Rewrite [Exponential Rule - Root Rewrite]:                               [tex]\displaystyle x^{\frac{1}{2}} + y^{\frac{1}{2}} = 2[/tex]
  2. [Function] Basic Power Rule:                                                                       [tex]\displaystyle \frac{1}{2}x^{\frac{1}{2} - 1} + \frac{1}{2}y^{\frac{1}{2} - 1}\frac{dy}{dx} = 0[/tex]
  3. [Derivative] Simplify:                                                                                     [tex]\displaystyle \frac{1}{2}x^{\frac{-1}{2}} + \frac{1}{2}y^{\frac{-1}{2}}\frac{dy}{dx} = 0[/tex]
  4. [Derivative] Rewrite [Exponential Rule - Rewrite]:                                       [tex]\displaystyle \frac{1}{2x^{\frac{1}{2}}} + \frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = 0[/tex]
  5. [Derivative] Isolate [tex]\displaystyle \frac{dy}{dx}[/tex] term [Subtraction Property of Equality]:                 [tex]\displaystyle \frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = -\frac{1}{2x^{\frac{1}{2}}}[/tex]
  6. [Derivative] Isolate [tex]\displaystyle \frac{dy}{dx}[/tex] [Multiplication Property of Equality]:                       [tex]\displaystyle \frac{dy}{dx} = -\frac{2y^{\frac{1}{2}}}{2x^{\frac{1}{2}}}[/tex]
  7. [Derivative] Simplify:                                                                                     [tex]\displaystyle \frac{dy}{dx} = -\frac{y^{\frac{1}{2}}}{x^{\frac{1}{2}}}[/tex]

Step 3: Evaluate

Find slope of tangent line

  1. Substitute in point [Derivative]:                                                                     [tex]\displaystyle \frac{dy}{dx} \bigg| \limit_{(\frac{9}{4}, \frac{1}{4})} = -\frac{(\frac{1}{4})^{\frac{1}{2}}}{(\frac{9}{4})^{\frac{1}{2}}}[/tex]
  2. [Slope] Exponents:                                                                                         [tex]\displaystyle \frac{dy}{dx} \bigg| \limit_{(\frac{9}{4}, \frac{1}{4})} = -\frac{\frac{1}{2}}{\frac{3}{2}}[/tex]
  3. [Slope] Simplify:                                                                                             [tex]\displaystyle \frac{dy}{dx} \bigg| \limit_{(\frac{9}{4}, \frac{1}{4})} = -\frac{1}{3}[/tex]

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

Unit: Differentiation - Implicit Differentiation

Book: College Calculus 10e

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