Mathematics, 11.04.2021 09:30 katyvillanueva05

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 poin

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Answer from: golderhadashaowtatz

B. $\displaystyle -\frac{1}{3}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to Right

Equality Properties

Multiplication Property of EqualityDivision Property of EqualityAddition Property of EqualitySubtraction Property of Equality

Algebra I

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

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) = cxⁿf’(x) = c·nxⁿ⁻¹

Implicit Differentiation

Step-by-step explanation:

Step 1: Define

$\displaystyle \sqrt{x} + \sqrt{y} = 2$

$\displaystyle (\frac{9}{4}, \frac{1}{4})$

Step 2: Differentiate

Implicit Differentiation

[Function] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle x^{\frac{1}{2}} + y^{\frac{1}{2}} = 2$ [Function] Basic Power Rule: $\displaystyle \frac{1}{2}x^{\frac{1}{2} - 1} + \frac{1}{2}y^{\frac{1}{2} - 1}\frac{dy}{dx} = 0$ [Derivative] Simplify: $\displaystyle \frac{1}{2}x^{\frac{-1}{2}} + \frac{1}{2}y^{\frac{-1}{2}}\frac{dy}{dx} = 0$ [Derivative] Rewrite [Exponential Rule - Rewrite]: $\displaystyle \frac{1}{2x^{\frac{1}{2}}} + \frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = 0$ [Derivative] Isolate $\displaystyle \frac{dy}{dx}$ term [Subtraction Property of Equality]: $\displaystyle \frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = -\frac{1}{2x^{\frac{1}{2}}}$ [Derivative] Isolate $\displaystyle \frac{dy}{dx}$ [Multiplication Property of Equality]: $\displaystyle \frac{dy}{dx} = -\frac{2y^{\frac{1}{2}}}{2x^{\frac{1}{2}}}$ [Derivative] Simplify: $\displaystyle \frac{dy}{dx} = -\frac{y^{\frac{1}{2}}}{x^{\frac{1}{2}}}$

Step 3: Evaluate

Find slope of tangent line

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

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

Unit: Differentiation - Implicit Differentiation

Book: College Calculus 10e

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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...