Respuesta :

TheHero
sin [sin(sinx)] 

You have two ways into this the first is:

sin y 

where y = sinz   and z = sinx

d(siny)/dx = cosy * dy/dx 

dy/dx = cosz * dz/dx

dz/dz = cosx 

and do reverse sub

dy/dz = cosz * cosx

d(siny)/dx = cosz * cosx * cosy  = cos(sin(sinx)) * cosx * cos(sinx)

or the way I prefer is take the derivative of each term in row:

cos(sin(sinx)) * cos(sinx) * cosx


Space

Answer:

[tex]\displaystyle \frac{dy}{dx} = \cos x \cos (\sin x) \cos \Big( \sin (\sin x) \Big)[/tex]

General Formulas and Concepts:

Calculus

Differentiation

  • Derivatives
  • Derivative Notation

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

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle y = \sin \Big( \sin (\sin x) \Big)[/tex]

Step 2: Differentiate

  1. Trigonometric Differentiation [Derivative Rule - Chain Rule]:                   [tex]\displaystyle y' = \cos \Big( \sin (\sin x) \Big) \Big( \sin (\sin x) \Big)'[/tex]
  2. Trigonometric Differentiation [Derivative Rule - Chain Rule]:                   [tex]\displaystyle y' = \cos (\sin x) \cos \Big( \sin (\sin x) \Big)(\sin x)'[/tex]
  3. Trigonometric Differentiation:                                                                       [tex]\displaystyle y' = \cos x \cos (\sin x) \cos \Big( \sin (\sin x) \Big)[/tex]

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

Unit: Differentiation

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