Solve 16x^3y/-4xy^3*-2xy/x

$$\frac { 16 x ^ { 3 } y } { - 4 x y ^ { 3 } } \cdot \frac { - 2 x y } { x }$$

Evaluate

$$\frac{8x^{2}}{y}$$

Solution Steps

Cancel out 4xy in both numerator and denominator.

$$\frac{4x^{2}}{-y^{2}}\times \frac{-2xy}{x}$$

Cancel out -1 in both numerator and denominator.

$$\frac{-4x^{2}}{y^{2}}\times \frac{-2xy}{x}$$

Cancel out x in both numerator and denominator.

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

Express \frac{-4x^{2}}{y^{2}}\left(-2\right) as a single fraction.

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

Express \frac{-\left(-4\right)x^{2}\times 2}{y^{2}}y as a single fraction.

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

Cancel out y in both numerator and denominator.

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

Multiply -4 and -1 to get 4.

$$\frac{4\times 2x^{2}}{y}$$

Multiply 4 and 2 to get 8.

$$\frac{8x^{2}}{y}$$

Differentiate w.r.t. x

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