calculus

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

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  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 ...

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

  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

  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.

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