Question

Choose an equivalent statement for f prime of pi over n if f of x equals quantity 1 minus cosine x close quantity over sine x.

1) f prime of x equals quantity 1 plus sine x close quantity over cosine x
2) f prime of x equals quantity 1 minus sine x close quantity over cosine x
3) f prime of x equals quantity 1 plus cosine x close quantity over sine x
4) f prime of x equals quantity 1 minus cosine x close quantity over sine x

Answer 1

After applying the quotient rule to the function f(x) = (1 - cos(x)) / sin(x), the derivative f'(x) does not match any of the answer choices given. This suggests there may be an error in the options provided for the equivalent statement of f prime of pi over n.

Step-by-step explanation:

The student has asked to choose an equivalent statement for f prime of pi over n if f of x equals quantity 1 minus cosine x close quantity over sine x. To find this, we must first determine the derivative, f prime of x, of the given function f(x) = (1 - cos(x)) / sin(x). This requires using the quotient rule of differentiation which is given by:

Quotient Rule: If g(x) and h(x) are differentiable functions, then the derivative of the quotient f(x) = g(x)/h(x) is given by

f'(x) = (g'(x)h(x) - g(x)h'(x)) / (h(x))^2

In our case, g(x) = 1 - cos(x) and h(x) = sin(x). Computing the derivatives g'(x) and h'(x), we get:

g'(x) = 0 + sin(x) and h'(x) = cos(x).

Plugging these into the quotient rule, we get:

f'(x) = (sin(x)sin(x) - (1 - cos(x))cos(x)) / sin^2(x)

Simplifying the numerator:

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

Given that sin^2(x) + cos^2(x) = 1, we can rewrite the numerator as:

f'(x) = (1 - cos(x)) / sin^2(x)

Which is the derivative of f(x). Looking at the answer choices, none of them exactly match the derivative we found. Hence, we can conclude that there may have been a mistake in the options provided, or alternatively, that the question may have been misinterpreted.