Find dy/dx: y= sqrt. 6/x^6 - sin2x

i know the first step is simplifying it to (6x^-6 - sin2x)^1/2. please show all steps.

so this is a chain rule problem...
you have simplified the expression

y = (6x^-6 - sin2x)^1/2

the first step is just to pretend that (6x^-6 - sin2x)^1/2 is "like" x^1/2,
so if you take the derivative of that,

1/2. (6x^-6 - sin2x)^-1/2

then, because of the chain rule, you also need to multiply by the derivative of the inside of the parentheses, which is -36x^-7 - 2cos 2x, so the answer is

1/2 (-36x^-7 - 2cos 2x) . (6x^-6 - sin2x)^-1/2

hope that helps :D

d(6x^-6 - sin2x)^1/2/dx let 6*x^-6 - sin2x=u

y=u^1/2

du/dx= -6*6*x^(-6-1) -2cos2x=
-36/x^7 -2cos2x

dy/du=1/2*u^(1/2-1)
=1/(2u^1/2)


dy/dx= dy/du*du/dx

1/(2u^1/2) * -36/x^7 - 2 cos2x

replace u

(-36-2x^7cos2x) / [ 2x^7(6 / x^6 - sin2x)^1/2)]

(-18-x^7cos2x) / [ x^7 (6/x^6- sin2x)^1/2)]