The Taylor expansion of a function f(x) about a point a can be written as:

f(x) = f(a) + f'(a)(x - a)/1! + f''(a)(x - a)^2/2! + f'''(a)(x - a)^3/3! + ...

In this case, we need to find the coefficient of x^3 in the Taylor expansion of sin(sin x) around x = 0.

Step 1: Find the derivatives
To find the Taylor expansion, we first need to find the derivatives of the function sin(sin x). Let's start by finding the first few derivatives:

f(x) = sin(sin x)
f'(x) = cos(sin x) * cos x
f''(x) = (cos^2(sin x) - sin^2(sin x)) * cos x - sin(sin x) * sin x
f'''(x) = (2sin(sin x)cos^3 x - 2sin^3(sin x)cos x - sin(sin x)) * cos x - (cos^2(sin x) - sin^2(sin x)) * sin x

Step 2: Evaluate the derivatives at x = 0
To find the Taylor expansion around x = 0, we need to evaluate the derivatives at x = 0. Let's calculate the values:

f(0) = sin(sin 0) = sin(0) = 0
f'(0) = cos(sin 0) * cos 0 = cos(0) * 1 = 1
f''(0) = (cos^2(sin 0) - sin^2(sin 0)) * cos 0 - sin(sin 0) * sin 0 = (1 - 0) * 1 - 0 = 1
f'''(0) = (2sin(sin 0)cos^3 0 - 2sin^3(sin 0)cos 0 - sin(sin 0)) * cos 0 - (cos^2(sin 0) - sin^2(sin 0)) * sin 0 = (0 - 0 - 0) * 1 - (1 - 0) * 0 = 0

Step 3: Write the Taylor expansion
Now that we have the values of the derivatives at x = 0, we can write the Taylor expansion up to the term with x^3:

f(x) = f(0) + f'(0)(x - 0)/1! + f''(0)(x - 0)^2/2! + f'''(0)(x - 0)^3/3!
= 0 + 1(x)/1! + 1(x^2)/2! + 0(x^3)/3!
= x + x^2/2

Step 4: Find the coefficient of x^3
The coefficient of x^3 is the term multiplied by x^3 in the Taylor expansion. In this case, the coefficient is 0. Therefore, the coefficient of x^3 in the Taylor expansion of sin(sin x) around x = 0 is 0.

The correct answer provided, between -0.35 and -0.30, does not match the calculated coefficient of 0. This suggests that there may be an error in either