Question

Chain Rule : Calculus problem
Find the derivative of
f(x) = sqrt[cos(5x)]
The answer is supposedly -5sin(5x)/2sqrtcos(5x)
How does one arrive at this answer?

Answer 1

Answer:

$\displaystyle f'(x) = \frac{-5sin(5x)}{2\sqrt{cos(5x)}}$

General Formulas and Concepts:

Algebra I

Functions

• Function Notation

Exponential Rule [Rewrite]:                                                                                      $\displaystyle b^{-m} = \frac{1}{b^m}$

Exponential Rule [Root Rewrite]:                                                                             $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Differentiation

• Derivatives
• Derivative Notation

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

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

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

Trig Derivatives

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = \sqrt{cos(5x)}$

Step 2: Differentiate

1. Rewrite [Exponential Rule - Root Rewrite]:                                                     $\displaystyle f(x) = [cos(5x)]^\bigg{\frac{1}{2}}$
2. Derivative Rule [Chain Rule]:                                                                           $\displaystyle f'(x) = \frac{d}{dx} \bigg[ [cos(5x)]^\bigg{\frac{1}{2}} \bigg] \cdot \frac{d}{dx}[cos(5x)] \cdot \frac{d}{dx}[5x]$
3. Rewrite [Derivative Property - Multiplied Constant]:                                      $\displaystyle f'(x) = \frac{d}{dx} \bigg[ [cos(5x)]^\bigg{\frac{1}{2}} \bigg] \cdot \frac{d}{dx}[cos(5x)] \cdot 5\frac{d}{dx}[x]$
4. Basic Power Rule:                                                                                            $\displaystyle f'(x) = \frac{1}{2}[cos(5x)]^\bigg{\frac{1}{2} - 1} \cdot \frac{d}{dx}[cos(5x)] \cdot 5x^{1 - 1}$
5. Simplify:                                                                                                             $\displaystyle f'(x) = \frac{5}{2}[cos(5x)]^\bigg{\frac{-1}{2}} \cdot \frac{d}{dx}[cos(5x)]$
6. Trig Derivative:                                                                                                 $\displaystyle f'(x) = \frac{5}{2}[cos(5x)]^\bigg{\frac{-1}{2}} \cdot -sin(5x)$
7. Simplify:                                                                                                             $\displaystyle f'(x) = \frac{-5sin(5x)}{2}[cos(5x)]^\bigg{\frac{-1}{2}}$
8. Rewrite [Exponential Rule - Rewrite]:                                                              $\displaystyle f'(x) = \frac{-5sin(5x)}{2[cos(5x)]^\bigg{\frac{1}{2}}}$
9. Rewrite [Exponential Rule - Root Rewrite]:                                                     $\displaystyle f'(x) = \frac{-5sin(5x)}{2\sqrt{cos(5x)}}$

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

Unit: Derivatives

Book: College Calculus 10e