Can somebody so I how to d/dx((6x^-6)/(1+4x^6)^3/2)

Answer 1

sre. Just use the quotient rule. If
y = u/v where u and v are functions of x, then
y' = (u'v - vu')/v^2
So, in this case, that gives
((-36x^-7)(1+4x^6)^(3/2) - (6x^-6)(3/2 (1+4x^6)^(1/2) (24x^5))/(1+4x^6)^3
You may want to simplify that a bit ...

Answer 2

oops. That is, of course, (u'v - uv')/v^2

Answer 3

(bottom* derivative of top - top * derivative of bottom) / bottom^2
bottom * derivative of top = (1+4x^6)^3/2) (-36 x^-7)
-top*derivative of bottom = -(6x^-6)(3/2)(1+4x^6)^1/2 *(24x^5)
bottom^2 = (1+4x^6)^3
plug and chug

Answer 4

or, if it looks a bit less messy, you can use the product rule. In that case, you have
y = (6/x^6)*(1+4x^6)^(-3/2)
y' = -36/x^7 * (1+4x^6)^(-3/2) + (6/x^6)(-3/2)(1+4x^6)^(-5/2) *(24x^5)
= -36/x^7 (1+4x^6)^(-5/2) (1+19x^6)
check my algebra, of course.