contestada

APPLICATION
3. Use the first principle of derivatives to determine the equation of tangent to the
curve y = 1/radical7x at
x = 5

APPLICATION 3 Use the first principle of derivatives to determine the equation of tangent to the curve y 1radical7x at x 5 class=

Respuesta :

Space

Answer:

[tex]\displaystyle y - \frac{\sqrt{35}}{35} = \frac{-\sqrt{35}}{350}(x - 5)[/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

Algebra I

Coordinates (x, y)

Point-Slope Form: y - y₁ = m(x - x₁)

  • x₁ - x coordinate
  • y₁ - y coordinate
  • m - slope

Algebra II

  • Exponential Rule [Rewrite]: [tex]\displaystyle b^{-m} = \frac{1}{b^m}[/tex]

Calculus

Derivatives

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

Derivative Notation

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Chain Rule: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]

Step-by-step explanation:

Step 1: Define

[tex]\displaystyle y = \frac{1}{\sqrt{7x}} \\x = 5[/tex]

Step 2: Differentiate

  1. [Function] Rewrite:                                                                                         [tex]\displaystyle y = \frac{1}{(7x)^\frac{1}{2}}[/tex]
  2. [Function] Rewrite [Exponential Rule - Rewrite]:                                         [tex]\displaystyle y = (7x)^{-\frac{1}{2}}[/tex]
  3. [Derivative] Chain Rule [Function]:                                                               [tex]\displaystyle y' = \frac{-1}{2}(7x)^{-\frac{1}{2} - 1} \cdot \frac{d}{dx}[7x][/tex]
  4. [Derivative] Simplify:                                                                                     [tex]\displaystyle y' = \frac{-1}{2}(7x)^{-\frac{3}{2}} \cdot \frac{d}{dx}[7x][/tex]
  5. [Derivative] Basic Power Rule:                                                                     [tex]\displaystyle y' = \frac{-1}{2}(7x)^{-\frac{3}{2}} \cdot 1 \cdot 7x^{1 - 1}[/tex]
  6. [Derivative] Simplify:                                                                                     [tex]\displaystyle y' = \frac{-7}{2}(7x)^{-\frac{3}{2}}[/tex]
  7. [Derivative] Rewrite [Exponential Rule - Rewrite]:                                       [tex]\displaystyle y' = \frac{-7}{2(7x)^{\frac{3}{2}}}[/tex]

Step 3: Find Slope of Tangent Line

  1. Substitute in x [Derivative]:                                                                           [tex]\displaystyle y'(5) = \frac{-7}{2[7(5)]^{\frac{3}{2}}}[/tex]
  2. [Tangent Slope] [Brackets] Multiply:                                                             [tex]\displaystyle y'(5) = \frac{-7}{2[35]^{\frac{3}{2}}}[/tex]
  3. [Tangent Slope] Evaluate exponents:                                                         [tex]\displaystyle y'(5) = \frac{-7}{2(35\sqrt{35})}[/tex]
  4. [Tangent Slope] Multiply:                                                                             [tex]\displaystyle y'(5) = \frac{-7}{70\sqrt{35}}[/tex]
  5. [Tangent Slope] Simplify:                                                                             [tex]\displaystyle y'(5) = \frac{-1}{10\sqrt{35}}[/tex]
  6. [Tangent Slope] Rationalize:                                                                         [tex]\displaystyle y'(5) = \frac{-\sqrt{35}}{350}[/tex]

Step 4: Find Tangent Line Equation

Point (x, y)

  1. Substitute in x [Function]:                                                                             [tex]\displaystyle y(5) = \frac{1}{\sqrt{7(5)}}[/tex]
  2. [Function] [√Radical] Multiply:                                                                     [tex]\displaystyle y(5) = \frac{1}{\sqrt{35}}[/tex]
  3. [Function] Rationalize:                                                                                   [tex]\displaystyle y(5) = \frac{\sqrt{35}}{35}[/tex]
  4. Define point:                                                                                                 [tex]\displaystyle (5, \frac{\sqrt{35}}{35})[/tex]

Equation

  1. Substitute in variables [Point-Slope Form]:                                                 [tex]\displaystyle y - \frac{\sqrt{35}}{35} = \frac{-\sqrt{35}}{350}(x - 5)[/tex]

Topic: AP Calculus AB/BC

Unit: Derivatives

Book: College Calculus 10e

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