Question

what is the derivative of y= 1/(2sin2x)

Answer 1

try using the quotient rule or the chain rule

Answer 2

\(\bf \cfrac{1}{2sin(2x)}\implies \cfrac{1}{2}\cdot [sin(2x)]^{-1}\) <-- chain rule it

Answer 3

so it would be - csc2x cot2x?

Answer 4

I got \(\bf \cfrac{1}{2sin(2x)}\implies \cfrac{1}{2}\cdot [sin(2x)]^{-1}
\\ \quad \\ \quad \\
-\cfrac{1}{2}\cdot [cos(2x)]^{-2}\cdot 2\)

Answer 5

i thought since the sin 2x would turn into csc 2x because thats the inverse of sin

Answer 6

hmm well, the notation for the inverse is \(\bf ^{-1}\) yes
however what we're taking is not the inverse function, but the reciprocal

Answer 7

ok so with the answer you got, it would translate to \[-\csc^22x \]