Solve -9x^3y*6xyz*1/3z^2y=

$$- 9 x ^ { 3 } y \cdot 6 x y z \cdot \frac { 1 } { 3 } z ^ { 2 } y =$$

Evaluate

$$-18x^{4}\left(yz\right)^{3}$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 3 and 1 to get 4.

$$-9x^{4}y\times 6yz\times \frac{1}{3}z^{2}y$$

Multiply y and y to get y^{2}.

$$-9x^{4}y^{2}\times 6z\times \frac{1}{3}z^{2}y$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$-9x^{4}y^{3}\times 6z\times \frac{1}{3}z^{2}$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$-9x^{4}y^{3}\times 6z^{3}\times \frac{1}{3}$$

Multiply -9 and 6 to get -54.

$$-54x^{4}y^{3}z^{3}\times \frac{1}{3}$$

Multiply -54 and \frac{1}{3} to get -18.

$$-18x^{4}y^{3}z^{3}$$

Differentiate w.r.t. x

$$-72\left(xyz\right)^{3}$$