Respuesta :

Hello from MrBillDoesMath!

Answer:

y' = -1/(1+sinx)

the fourth choice.



Discussion:

Using the derivative quotient rule

y' =    (  (1 + sinx) (-sinx) - cosx ( cosx)  )  \  ( 1 + sinx)^2

as (cosx)' = -sinx and  (sinx)' = cos

Expanding we get

y' =     (  -sinx - (sinx)^2 - (cosx)^2 ) \ (1 + sinx)^2

But (sinx)^2 + (cos)^2 = 1 so this equals

y' = ( -sinx -1 )\ ( 1 + sinx)^2 =>

y' = - ( 1 + sinx) / (1 + sinx)^2

Notice the numerator is the square of the denominator so

y' = -1/(1+sinx)


which is the fourth choice



Thank you,

MrB


Space

Answer:

[tex]\displaystyle y' = \frac{-1}{1 + \sin x}[/tex]

General Formulas and Concepts:

Calculus

Differentiation

  • Derivatives
  • Derivative Notation

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

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

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

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle y = \frac{\cos x}{1 + \sin x}[/tex]

Step 2: Differentiate

  1. [Function] Derivative Rule [Quotient Rule]:                                                 [tex]\displaystyle y' = \frac{(\cos x)'(1 + \sin x) - \cos x(1 + \sin x)'}{(1 + \sin x)^2}[/tex]
  2. Trigonometric Differentiation:                                                                       [tex]\displaystyle y' = \frac{-\sin x(1 + \sin x) - \cos x(1 + \sin x)'}{(1 + \sin x)^2}[/tex]
  3. Rewrite [Derivative Rule - Addition/Subtraction]:                                       [tex]\displaystyle y' = \frac{-\sin x(1 + \sin x) - \cos x[(1)' + (\sin x)']}{(1 + \sin x)^2}[/tex]
  4. Trigonometric Differentiation:                                                                       [tex]\displaystyle y' = \frac{-\sin x(1 + \sin x) - \cos x[(1)' + \cos x]}{(1 + \sin x)^2}[/tex]
  5. Basic Power Rule:                                                                                         [tex]\displaystyle y' = \frac{-\sin x(1 + \sin x) - \cos^2 x}{(1 + \sin x)^2}[/tex]
  6. Factor:                                                                                                           [tex]\displaystyle y' = \frac{- \big[ \sin x(1 + \sin x) + \cos^2 x \big]}{(1 + \sin x)^2}[/tex]
  7. Expand:                                                                                                         [tex]\displaystyle y' = \frac{- \big[ \sin x+ \sin^2 x + \cos^2 x \big]}{(1 + \sin x)^2}[/tex]
  8. Simplify:                                                                                                         [tex]\displaystyle y' = \frac{-(\sin x + 1)}{(1 + \sin x)^2}[/tex]
  9. Simplify:                                                                                                         [tex]\displaystyle y' = \frac{-1}{1 + \sin x}[/tex]

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

Unit: Differentiation

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