Solve (2x^{-3}y^{4})^{3}(x^3+y)^0/(4xy^-2)^3

$$( 2 x ^ { - 3 } y ^ { 4 } ) ^ { 3 } ( x ^ { 3 } + y ) ^ { 0 } / ( 4 x y ^ { - 2 } ) ^ { 3 }$$

Evaluate

$$\frac{y^{18}}{8x^{12}}$$

Solution Steps

Expand \left(2x^{-3}y^{4}\right)^{3}.

$$\frac{2^{3}\left(x^{-3}\right)^{3}\left(y^{4}\right)^{3}\left(x^{3}+y\right)^{0}}{\left(4xy^{-2}\right)^{3}}$$

To raise a power to another power, multiply the exponents. Multiply -3 and 3 to get -9.

$$\frac{2^{3}x^{-9}\left(y^{4}\right)^{3}\left(x^{3}+y\right)^{0}}{\left(4xy^{-2}\right)^{3}}$$

To raise a power to another power, multiply the exponents. Multiply 4 and 3 to get 12.

$$\frac{2^{3}x^{-9}y^{12}\left(x^{3}+y\right)^{0}}{\left(4xy^{-2}\right)^{3}}$$

Calculate 2 to the power of 3 and get 8.

$$\frac{8x^{-9}y^{12}\left(x^{3}+y\right)^{0}}{\left(4xy^{-2}\right)^{3}}$$

Calculate x^{3}+y to the power of 0 and get 1.

$$\frac{8x^{-9}y^{12}\times 1}{\left(4xy^{-2}\right)^{3}}$$

Multiply 8 and 1 to get 8.

$$\frac{8x^{-9}y^{12}}{\left(4xy^{-2}\right)^{3}}$$

Expand \left(4xy^{-2}\right)^{3}.

$$\frac{8x^{-9}y^{12}}{4^{3}x^{3}\left(y^{-2}\right)^{3}}$$

To raise a power to another power, multiply the exponents. Multiply -2 and 3 to get -6.

$$\frac{8x^{-9}y^{12}}{4^{3}x^{3}y^{-6}}$$

Calculate 4 to the power of 3 and get 64.

$$\frac{8x^{-9}y^{12}}{64x^{3}y^{-6}}$$

Cancel out 8 in both numerator and denominator.

$$\frac{x^{-9}y^{12}}{8y^{-6}x^{3}}$$

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent.

$$\frac{x^{-9}y^{18}}{8x^{3}}$$

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

$$\frac{y^{18}}{8x^{12}}$$

Differentiate w.r.t. x

$$-\frac{3y^{18}}{2x^{13}}$$