Math help plzzzz?

Question

Differenciate y=x^2(x-1)(x^2+4)^3

Answer 1

ln y = ln x² (x-1)(x²+4)³
ln y = ln x² + ln (x-1) + ln(x²+4)³
ln y = 2 ln x + ln (x-1) + 3 ln (x²+4)

y'/y = ( 2 / y ) + ( 1 / x-1 ) + 3 * ( 2x / x²+4 )

y' = y [ ( 2 / x ) + ( 1 / x-1 ) + ( 6x / x²+4 ) ]
y' = x² ( x - 1 ) ( x²+4 ) [ ( 2 / x ) + ( 1 / x-1 ) + ( 6x / x²+4 ) ]

Answer 2

Use the product rule:

y' = (x-1)(x^2+4)^3(2x) + x^2(x^2+4)^3 + 3x^2(x-1)(x^2+4)^2(2x)

Answer 3

y = x²(x-1)(x²+4)³
y = (x³-x²) (x² + 4)³
y' = (x³-x²) * 3 (x² + 4)² * 2x + (x² + 4)³ (3x² - 2x)
y' = 6x³ (x-1) (x² + 4)² + x(x² + 4)³ (3x- 2)
y' = x(x² + 4)² (6x² (x-1) + (x² + 4) (3x- 2))

Answer 4

y=x^2(x-1)(x^2+4)^3
dy/dx=d/dx [x^2(x-1)(x^2+4)^3]
dy/dx = x^2(x-1) d/dx[(x^2+4)^3] + (x^2+4)^3 d/dx [x^2(x-1)]
dy/dx = x^2(x-1) 3[(x^2+4)^3-1] d/dx(x^2+4) + (x^2+4)^3 [d/dx [(x^3-x^2)]
dy/dx = 3x^2(x-1) (x^2+4)^2 (2x) + (x^2+4)^3 (3x^2-2x)
dy/dx = 6x^3(x-1) (x^2+4)^2 + (x^2+4)^3 (3x^2-2x)
dy/dx = 6x^3(x-1) (x^2+4)^2 + x(x^2+4)^3 (3x-2)

then factor out x(x^2+4)^2
dy/dx =x(x^2+4)^2 [ 6x^2(x-1) + (x^2+4)(3x-2)]
dy/dx =x(x^2+4)^2 [ 6x^3 - 6x^2 + 3x^3+12x -2x^2 - 8]
dy/dx =x(x^2+4)^2 [ 9x^3 - 8x^2 +12x - 8]