derivative of the function

[jon12]

I worked this out and this is what I got, does it look right.

y = squareroot(3x-7)(x^4-1)^3

y' = (1/2)(3)(x^4-1)^3/sqrt(3x-7) +(3).sqrt(3x-7)(x^4-1)^2*(4x^3)

y' = (x^4-1)^2[(3/2)(x^4-1) +12x^3(3x-7)] / sqrt(3x-7)

y' = 3(x^4-1)^2[x^4-1 +24x^4-56] / 2*sqrt(3x-7)

y' = 3(x^4-1)^2(25x^4 - 57) / 2*sqrt(3x-7)

 

[Deleted member 4993]

I worked this out and this is what I got, does it look right.

y = squareroot(3x-7)(x^4-1)^3

y' = (1/2)(3)(x^4-1)^3/sqrt(3x-7) +(3).sqrt(3x-7)(x^4-1)^2*(4x^3)

y' = (x^4-1)^2[(3/2)(x^4-1) +12x^3(3x-7)] / sqrt(3x-7)

y' = 3(x^4-1)^2[x^4-1 +24x^4-56x^3] / 2*sqrt(3x-7)

y' = 3(x^4-1)^2(25x^4 - 56x^3 - 1) / 2*sqrt(3x-7)

Correct - except as (corrected) shown in red.