How do I use quotient rule to show (d/dx)(cot(x)) = -cosec(x)?

Question

How do I use quotient rule to show (d/dx)(cot(x)) = -cosec(x)

Answer 1

cot(x) can be written as cos(x) / sin(x)
Now apply the quotient rule
d/dx (cotx)
= d/dx (cosx / sinx)
= [(sinx)(-sinx) - (cosx)(cosx)] / (sinx)^2
= [-(sinx)^2 - (cosx)^2] / (sinx)^2
= (-1)[(sinx)^2 + (cosx)^2] / (sinx)^2
= (-1)[1] / (sinx)^2
= -1/(sinx)^2
= -(cscx)^2

Answer 2

There is a trigonometric property that says cot(x) = cos(x)/sin(x).

When you writ it in this form, you now have a "top" and "bottom" function so you can use the quotient rule.

You will get (d/dx)(cos(x)/sin(x) = [(cos(x))'sin(x) - cos(x)(sin(x))']/sin²(x)

= [(-sin(x))sin(x) - cos(x)cos(x)]/sin²(x)

= [-sin²(x) - cos²(x)]/sin²(x)

= -1/sin²(x)

= -cosec²(x)

Be careful as the answer is cosec²(x) not cosec(x).

Answer 3

cot(x) = cos(x)/sin(x)

d/dx (cot(x)) = (sin(x)*d/dx(cos(x)) - cos(x)*d/dx(sin(x))) / sin^2(x)
= (sin(x)*(-sin(x)) - cos(x)*cos(x)) / sin^2(x)
= (-sin^2(x) - cos^2(x)) / sin^2(x)
= -1 / sin^2(x)
= -csc^2(x)

The answer is NOT -csc(x).

Answer 4

Find (d/dx)(cot(x)).

cot(x) = cos(x)/sin(x)

(d/dx)(cot(x)) = (d/dx){cos(x)/sin(x)}
= {sin(x)[(d/dx)(cos(x))] - cos(x)[(d/dx)(sin(x))]}/sin²(x)
= {(sin(x))(-sin(x)) - (cos(x))(cos(x))}/sin²(x)
= {-sin²(x) - cos²(x)}/sin²(x)
= -1/sin²(x) = -csc²(x)