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

APPLICATION3. Use The First Principle Of Derivatives To Determine The Equation Of Tangent To Thecurve

Answers

Answer 1

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

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to Right

Algebra I

Coordinates (x, y)

Point-Slope Form: y - y₁ = m(x - x₁)

x₁ - x coordinatey₁ - y coordinatem - 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) = cxⁿf’(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

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

Step 3: Find Slope of Tangent Line

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

Step 4: Find Tangent Line Equation

Point (x, y)

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

Equation

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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