Question

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)

Answer 1

Answer:

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

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]:                                                                           $\displaystyle b^{-m} = \frac{1}{b^m}$
• 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) = cxⁿ
• f’(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

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

Step 3: Evaluate

Find slope of tangent line

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