Derivative quotient rule

I need some help with steps, and if someone can explain why the denominator would be (3x-4)^3 and not 4.
Find derivative : [4x^7] / [3x - 4] ^3
Thanks

the derivative of (u / v) = [u ' v - v ' u] / (v^2)

here, u = 4x^7 ==> u' = 28x^6
v = (3x - 4)^3 ==> v ' = 3(3x - 4)^2 (3) = 9(3x - 4)^2
v^2 = (3x - 4)^6

plug in to give the derivative is:
[28x^6(3x - 4)^3 - 4x^7 (9(3x - 4)^2)] / [3x - 4]^6

factor (3x - 4)^2 out of the top to give:
derivative = (3x - 4)^2 [28x^6 (3x - 4) - 4x^7 (9)] / [(3x - 4)^6]
cancel and distribute to give:
[84x^7 - 112x^6 - 36x^7] / (3x - 4)^4

further simplify to : derivative = [48 x ^7 - 112 x^6] / (3x - 4)^4
or even further to 16x^6 (3x - 7) / (3x - 4)^4

[4x^7] / [3x - 4] ^3 = [4x^7] * [3x - 4] ^(-3) from just algebra

(d/dx)([4x^7] * [3x - 4] ^(-3)} = (d/dx)([4x^7] + (d/dx) [3x - 4] ^(-3)}

= (d/dx)([4x^7] + (d/dx) [3x - 4] ^(-3)} = 7*4(x^6) + (-3)*{[3x-4]^(-4)}*(d/dx)[3x-4]

= 28x^6 + (-3)*{[3x-4]^(-4)}*(3) = 28x^6 - 9*[3x-4]^(-4) = 28x^6 - 9/{[3x-4]^(4)}

The rest is just more algebra, which I will let you do for the practice.
I have always found it easier in finding a derivative to algebraically rewrite a fractional expression, such as the starting one here, as a product of the numerator expression and the negative power of the denominator expression, as I did above. You or your teacher may prefer using the "standard formula" for divided expressions.

The derivative is 16x^6(3x - 7)/(3x - 4)^4.