Need Math help for calculus.?

1) (16x^2 y)^3/4

= 16^3/4 * (x^2)^3/4 * y^3/4

= (2^4)^3/4 * x^3/2 * y^3/4

= 8x^3/2y^3/4

2) 2(2/2-x) * [-2/(2-x)^2]

= 4/(2 - x) * - 2/(2 - x)^2

= - 8/(2 - x)^3

Hope I have not wasted my time

1) (16x^2y)^(3/4)

When you have a base raised to a fractional power, keep in mind that the numerator is the integer power that the base is raised to, and the denominator is the integer ROOT of said base. Therefore, (16x^2y)^(3/4)...

= 4th root((16x^2y)^3)

= 4th root((2^4 * x^2 * y)^3)

= 4th root(2^(4 * 3) * x^(2 * 3) * y^(1 * 3))

= 4th root(2^12 * x^6 * y^3)

= 2^(12/4) * x^(6/4) * y^(3/4)

= 8x 4th root(x^2y^3).

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2) 2(2 / (2 - x)) * (-2 / (2 - x)^2)

This is nothing but multiplying with fractions. Just multiply the numerators by themselves, then the denominators. You need not find a least common denominator.

= 2(2)(-2) / ((2 - x) * (2 - x)^2)

When you have the same base multiple times in an expression, but it's raised to different powers, add those powers. In this case:

= 2(2)(-2) / (2 - x)^(1 + 2)

= -8 / (2 - x)^3

= -8 / (8 - 12x + 6x^2 - x^3). Note that x =/= 2 because of division by zero.

1)

As written:

(16x² y)³/4

= (16)³/4 * (x²)³ * y³

= 4096/4 * x^(²*³) * y³

= 1024 x⁶ y³

2)

NOT as written, but as I think you meant:

2(2/(2−x)) * [−2/(2−x)²]

= (2*2*−2)/[(2−x)(2−x)²]

= −8/(2−x)³

= 8/(x−2)³

rules

(ab)^m =a^m b^m

and

(a^m)^n=a^(mn)