Mathematics, 23.02.2021 05:30 katia87

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

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Answer from: DineroLaForeign

$\displaystyle y - \frac{\sqrt{35}}{35} = \frac{-\sqrt{35}}{350}(x - 5)$

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]: $\displaystyle b^{-m} = \frac{1}{b^m}$

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: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \frac{1}{\sqrt{7x}} \\x = 5$

Step 2: Differentiate

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

Step 3: Find Slope of Tangent Line

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

Step 4: Find Tangent Line Equation

Point (x, y)

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

Equation

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

Topic: AP Calculus AB/BC

Unit: Derivatives

Book: College Calculus 10e

Answer from: Quest

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Answer from: Quest

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step-by-step explanation:

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