Question

Derivative of [sin [(x+1)/(x-1)]]^3?
show steps please

Answer 1

you need to use the chain rule several times

like let u = (x + 1)/(x -1)

so you have sin^3(u)

then let v = sin u

and take the derivative of v^3

back substitute everything in

its long, but doable

Answer 2

I have 3[sin (x+1)/(x-1)]^2 times cos(x+10/(x-1)

Answer 3

not sure. try using wolframalpha to verify if that is correct. what you get should be equivalent to

\[-\frac{ 3 \sin \left( \frac{ x+1 }{ x - 1 } \sin \left( \frac{ 2(x+1) }{ x - 1 } \right) \right) }{ (x - 1)^{2} }\]