Question

What is the derivative of sin sin sin x

Answer 1

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

Answer 2

`\displaystyle (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 (d)/(dx)[f(g(x))] =f'(g(x)) \cdot g'(x)`

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