Solve frac{(2x^{-4}y^2)^-4}{x^0y^-1}=

$$\frac { ( 2 x ^ { - 4 } y ^ { 2 } ) ^ { - 4 } } { x ^ { 0 } y ^ { - 1 } } =$$

Evaluate

$$\frac{x^{16}}{16y^{7}}$$

Solution Steps

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

$$\frac{2^{-4}\left(x^{-4}\right)^{-4}\left(y^{2}\right)^{-4}}{x^{0}y^{-1}}$$

To raise a power to another power, multiply the exponents. Multiply -4 and -4 to get 16.

$$\frac{2^{-4}x^{16}\left(y^{2}\right)^{-4}}{x^{0}y^{-1}}$$

To raise a power to another power, multiply the exponents. Multiply 2 and -4 to get -8.

$$\frac{2^{-4}x^{16}y^{-8}}{x^{0}y^{-1}}$$

Calculate 2 to the power of -4 and get \frac{1}{16}.

$$\frac{\frac{1}{16}x^{16}y^{-8}}{x^{0}y^{-1}}$$

Calculate x to the power of 0 and get 1.

$$\frac{\frac{1}{16}x^{16}y^{-8}}{1y^{-1}}$$

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

$$\frac{\frac{1}{16}x^{16}}{y^{7}}$$

Differentiate w.r.t. x

$$\frac{x^{15}}{y^{7}}$$