Need help with this

Help with num 3 please. thanks

Alja [10]

Answer:

a) $\displaystyle \frac{dy}{dx} \bigg| \limits_{x = 0} = -1$

b) $\displaystyle \frac{dy}{dx} \bigg| \limits_{x = \frac{\pi}{2}} = -1$

General Formulas and Concepts:

Pre-Calculus

Unit Circle

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:

f(x) = cxⁿ
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 [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

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

Trigonometric Differentiation

Logarithmic Differentiation

Step-by-step explanation:

a)

Step 1: Define

Identify

$\displaystyle y = ln \bigg( \frac{1 - x}{\sqrt{1 + x^2}} \bigg)$

Step 2: Differentiate

Logarithmic Differentiation [Chain Rule]: $\displaystyle \frac{dy}{dx} = \frac{1}{\frac{1 - x}{\sqrt{1 + x^2}}} \cdot \frac{d}{dx}[\frac{1 - x}{\sqrt{1 + x^2}}]$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{d}{dx}[\frac{1 - x}{\sqrt{1 + x^2}}]$
Quotient Rule: $\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{(1 - x)'\sqrt{1 + x^2} - (1 - x)(\sqrt{1 + x^2})'}{(\sqrt{1 + x^2})^2}$
Basic Power Rule [Chain Rule]: $\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{-\sqrt{1 + x^2} - (1 - x)(\frac{x}{\sqrt{x^2 + 1}})}{(\sqrt{1 + x^2})^2}$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \bigg( \frac{x(x - 1)}{(x^2 + 1)^\bigg{\frac{3}{2}}} - \frac{1}{\sqrt{x^2 + 1}} \bigg)$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{x + 1}{(x - 1)(x^2 + 1)}$

Step 3: Find

Substitute in x = 0 [Derivative]: $\displaystyle \frac{dy}{dx} \bigg| \limit_{x = 0} = \frac{0 + 1}{(0 - 1)(0^2 + 1)}$
Evaluate: $\displaystyle \frac{dy}{dx} \bigg| \limits_{x = 0} = -1$

b)

Step 1: Define

Identify

$\displaystyle y = ln \bigg( \frac{1 + sinx}{1 - cosx} \bigg)$

Step 2: Differentiate

Logarithmic Differentiation [Chain Rule]: $\displaystyle \frac{dy}{dx} = \frac{1}{\frac{1 + sinx}{1 - cosx}} \cdot \frac{d}{dx}[\frac{1 + sinx}{1 - cosx}]$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{d}{dx}[\frac{1 + sinx}{1 - cosx}]$
Quotient Rule: $\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{(1 + sinx)'(1 - cosx) - (1 + sinx)(1 - cosx)'}{(1 - cosx)^2}$
Trigonometric Differentiation: $\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{cos(x)(1 - cosx) - sin(x)(1 + sinx)}{(1 - cosx)^2}$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - sin(x) - 1]}{[sin(x) + 1][cos(x) - 1]}$

Step 3: Find

Substitute in x = π/2 [Derivative]: $\displaystyle \frac{dy}{dx} \bigg| \limit_{x = \frac{\pi}{2}} = \frac{-[cos(\frac{\pi}{2}) - sin(\frac{\pi}{2}) - 1]}{[sin(\frac{\pi}{2}) + 1][cos(\frac{\pi}{2}) - 1]}$
Evaluate [Unit Circle]: $\displaystyle \frac{dy}{dx} \bigg| \limit_{x = \frac{\pi}{2}} = -1$

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

Unit: Differentiation

Book: College Calculus 10e

4 0

3 years ago

7 (x-4) please help

Rufina [12.5K]

Simplify. Distribute 7 to all terms within the parenthesis

7(x) = 7x

7(-4) = -28

7x - 28 is your answer

hope this helps

7 0

3 years ago

Read 2 more answers

The fraction 5/8 produces a repeating decimal. ___ 0.625 True or False

Rama09 [41]

Answer:

False i just did it

Step-by-step explanation:

7 0

3 years ago

Please answer this correctly

Leviafan [203]

Answer:

Skateboard:

Scooter:

Bike:

Step-by-step explanation:

19 + 14 + 17 = 50 (there were 50 customers)

19 out of 50 = 38%

14 out of 50 = 28%

17 out of 50 = 34%

Hope this helps!

5 0

4 years ago

Help me ok thx!!!!!!!!!!!!

MA_775_DIABLO [31]

Answer:

5 kelomiters

7 0

3 years ago

Need help with this​

Need help with this