Web Analytics

Verified Answer :

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:

\(\displaystyle \frac{dy}{dx} = \cos x \cos (\sin x) \cos \Big( \sin (\sin x) \Big)\)

General Formulas and Concepts:

Calculus

Differentiation

  • Derivatives
  • Derivative Notation

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

\(\displaystyle y = \sin \Big( \sin (\sin x) \Big)\)

Step 2: Differentiate

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

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

Unit: Differentiation

Related Questions