Question

Calculate f prime of pi over 3 for f of x equals cot x over sin x period

A. negative 10 times the square root of 3 over 9
B. negative 8 over 3


It is either A or B, the other two answers are incorrect.

Answer 1

Negative 8/3!!!!

Explanation:

To calculate f'(π/3) for f(x) = cot(x)/sin(x), we need to find the derivative of f(x) and then substitute x = π/3.

Let's begin by expressing f(x) in terms of sin and cos functions:

f(x) = cot(x)/sin(x) = cos(x)/sin(x) ÷ sin(x) = cos(x)/(sin(x))^2

Now, we can find the derivative of f(x) using the quotient rule:

f'(x) = (sin(x))^2 * (-sin(x)) - cos(x) * 2sin(x) * cos(x) / (sin(x))^4

Next, let's substitute x = π/3:

f'(π/3) = (sin(π/3))^2 * (-sin(π/3)) - cos(π/3) * 2sin(π/3) * cos(π/3) / (sin(π/3))^4

To simplify this expression, we can use the following trigonometric identities:

sin(π/3) = √3/2 cos(π/3) = 1/2

Substituting these values, we get:

f'(π/3) = (√3/2)^2 * (-(√3/2)) - (1/2) * 2 * (√3/2) * (1/2) / (√3/2)^4

Simplifying further:

f'(π/3) = (3/4) * (-(√3/2)) - (1/2) * (√3/2) * (1/2) / (3/4)^2

f'(π/3) = (-3√3/8) - (3/8) / 9/16

f'(π/3) = (-3√3 - 3) / 9 = - (3√3 + 3) / 9 = - (√3 + 1) / 3

Therefore, the value of f'(π/3) is option B. negative 8/3.

Nice