Solve (14pq^{2}r^3)(-10p^2q^4r)(-pqr^2)

$$( 14 p q ^ { 2 } r ^ { 3 } ) ( - 10 p ^ { 2 } q ^ { 4 } r ) ( - p q r ^ { 2 } )$$

Evaluate

$$140p^{4}r^{6}q^{7}$$

Solution Steps

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

$$14p^{3}q^{2}r^{3}\left(-10\right)q^{4}r\left(-p\right)qr^{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 4 to get 6.

$$14p^{3}q^{6}r^{3}\left(-10\right)r\left(-p\right)qr^{2}$$

To multiply powers of the same base, add their exponents. Add 6 and 1 to get 7.

$$14p^{3}q^{7}r^{3}\left(-10\right)r\left(-p\right)r^{2}$$

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

$$14p^{3}q^{7}r^{4}\left(-10\right)\left(-p\right)r^{2}$$

To multiply powers of the same base, add their exponents. Add 4 and 2 to get 6.

$$14p^{3}q^{7}r^{6}\left(-10\right)\left(-p\right)$$

Multiply 14 and -10 to get -140.

$$-140p^{3}q^{7}r^{6}\left(-p\right)$$

Multiply -140 and -1 to get 140.

$$140p^{3}q^{7}r^{6}p$$

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

$$140p^{4}q^{7}r^{6}$$

Differentiate w.r.t. p

$$560p^{3}r^{6}q^{7}$$