Question

If the gradient of the tangent to

Answer 1

Answer:

Point A(9, 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  

• Coordinates (x, y)
• Functions
• Function Notation
• Terms/Coefficients
• Anything to the 0th power is 1
• 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  

Basic Power Rule:  

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

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

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle y = \sqrt{x}$

$\displaystyle y' = \frac{1}{6}$

Step 2: Differentiate

1. [Function] Rewrite [Exponential Rule - Root Rewrite]:                                   $\displaystyle y = x^{\frac{1}{2}}$
2. Basic Power Rule:                                                                                             $\displaystyle y' = \frac{1}{2}x^{\frac{1}{2} - 1}$
3. Simplify:                                                                                                             $\displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}}$
4. [Derivative] Rewrite [Exponential Rule - Rewrite]:                                          $\displaystyle y' = \frac{1}{2x^{\frac{1}{2}}}$
5. [Derivative] Rewrite [Exponential Rule - Root Rewrite]:                                 $\displaystyle y' = \frac{1}{2\sqrt{x}}$

Step 3: Solve

Find coordinates of A.

x-coordinate

1. Substitute in y' [Derivative]:                                                                             $\displaystyle \frac{1}{6} = \frac{1}{2\sqrt{x}}$
2. [Multiplication Property of Equality] Multiply 2 on both sides:                      $\displaystyle \frac{1}{3} = \frac{1}{\sqrt{x}}$
3. [Multiplication Property of Equality] Cross-multiply:                                      $\displaystyle \sqrt{x} = 3$
4. [Equality Property] Square both sides:                                                           $\displaystyle x = 9$

y-coordinate

1. Substitute in x [Function]:                                                                                $\displaystyle y = \sqrt{9}$
2. [√Radical] Evaluate:                                                                                         $\displaystyle y = 3$

∴ Coordinates of A is (9, 3).

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

Unit: Derivatives

Book: College Calculus 10e