Question

Show that d/dx(csc(x)) = -csc(x) cot(x).

a) csc(x) cot(x)
b) -csc(x) cot(x)
c) cot(x)
d) -csc(x)

Answer 1

The derivative of csc(x) is -csc(x)cot(x), which is found using the quotient rule for derivatives and simplifying with trigonometric identities.

Step-by-step explanation:

To show that the derivative of csc(x) is -csc(x)cot(x), we will use the quotient rule and trigonometric identities. The cosecant function csc(x) is the reciprocal of the sine function, so csc(x) = 1/sin(x).

Step 1: Express csc(x) in terms of sine

csc(x) = 1/sin(x)

Step 2: Differentiate using the quotient rule

The quotient rule states that if you have a function g(x) that is the quotient of two functions, u(x) / v(x), then its derivative is given by (v(x)u'(x) - u(x)v'(x)) / v(x)^2.

Here, u(x) = 1 and v(x) = sin(x), so u'(x) = 0 and v'(x) = cos(x).

Applying the quotient rule gives us:

d/dx(csc(x)) = (sin(x) · 0 - 1 · cos(x)) / sin(x)^2 = -cos(x) / sin(x)^2

Step 3: Simplify using trigonometric identities

We can write cos(x)/sin(x) as cot(x) and 1/sin(x) as csc(x), so:

d/dx(csc(x)) = -cot(x)csc(x)

Therefore, the derivative of csc(x) is -csc(x)cot(x), which corresponds to option b.