Can you pls show me how to differentiate (cos(2x))/(sin(2x))

The answer is -2/(sin^2(2x))

thanks

f'(x) = [(-2sin2x)(sin2x) - (2cos2x)(cos2x)] / (sin2x)^2
f'(x) = (-2sin^2 2x - 2cos^2 2x) / sin^2 2x
f'(x) = (-2(1 - cos^2 2x) - 2cos^2 2x) / sin^2 2x
f'(x) = (-2 + 2co^2 2x - 2cos^2 2x) / sin^2 2x
f'(x) = -2 / sin^2(2x)

tan(x) = sin(x)/cos(x), so tan(2x) = sin(2x)/cos(2x)
in this case we are trying to derive:
d/dx [cos(2x)/sin(2x)]
= d/dx [1/tan(2x)]
from here you can either equate this to cot(2x) and derive using the fact that the derivative of cot(x) = -csc^2(x), or you can use the quotient rule (which is simplified by the fact that the numerator is 1)

= [tan(2x)*0 - 1*2sec^2(2x)]/[tan^2(2x)]
= -2sec^2(2x)/tan^2(2x)
= -2/cos^2(2x) * cos^2(2x)/sin^2(2x)
= -2/sin^2(2x)
= -2csc^2(2x)

which is the required answer

You'll need to use the quotient rule and the chain rule.

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

Chain rule:
f(g(x)) = f'(g(x)) * g'(x)


f(x) = cos(2x) / sin(2x)
Chain rule and quotient rule:
f'(x) = -2sin^2(2x) - 2cos^2(2x) / sin^2(2x)
Simplify with Pythagorean identity:
f'(x) = -2[sin^2(2x) + cos^2(2x)] / sin^2(2x)
f'(x) = -2 / sin^2(2x)

How I used the chain rule:
h(x) = cos(2x)
h'(x) = -sin(2x) * 2 = -2sin(2x)

g(x) = sin(2x)
g'(x) = cos(2x) * 2 = 2cos(2x)

Let y=(cos(2x))/(sin(2x))

Let u=cos(2x) and v=sin(2x)

du/dx=-2sin(2x) dv/dx=2cos(2x)

Using the quotient rule dy/dx=(v(du/dx)-u(dv/dx))/v^2

dy/dx=((sin(2x))(-2sin(2x))-(cos(2x))(…

=(-2sin^2(2x)-2cos^2(2x))/(sin^2(2x))

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

Since sin^2(a)+cos^2(a)=1 (That's a rule. a can be any angle as long as both a's are the same angle) Then
dy/dx=-2/(sin^2(2x))

Hope that helped.

Use the quocient rule.... (p/q)´ = (p´q -pq´)/q^2

As y = (cos(2x))/ (sin(2x)) then y´= [2(-sin(2x))*sin(2x) - (cos(2x))*2cos(2x)] /(sin(2x))^2

in the numerator put -2 in evidence and (cos(2x))^2 + (sin(2x))^2 = 1 then

y = -2/(sin(2x))^2 ... the answer is correct. OK!