Question

What is the derivative of x(sqrt (16-x^2)

Answer 1

Answer:

$\displaystyle \frac{dy}{dx} = \sqrt{16 - x^2} - \frac{x^2}{\sqrt{16 - x^2}}$

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Multiplied Constant]:                                                           $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]:                                                        $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$  

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

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

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

$\displaystyle y = x\sqrt{16 - x^2}$

Step 2: Differentiate

1. Derivative Rule [Product Rule]:                                                                     $\displaystyle y' = (x)'\sqrt{16 - x^2} + x(\sqrt{16 - x^2})'$
2. Basic Power Rule [Derivative Rule - Chain Rule]:                                       $\displaystyle y' = \sqrt{16 - x^2} + \frac{x}{2\sqrt{16 - x^2}}(16 - x^2)'$
3. Basic Power Rule [Derivative Properties]:                                                  $\displaystyle y' = \sqrt{16 - x^2} - \frac{x^2}{2\sqrt{16 - x^2}}$

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

Unit: Differentiation