Solve 9b/16b^2*32b^2h/-27h^2b^5

$$\frac { 9 b } { 16 b ^ { 2 } } \times \frac { 32 b ^ { 2 } h } { - 27 h ^ { 2 } b ^ { 5 } }$$

Evaluate

$$-\frac{2}{3hb^{4}}$$

Solution Steps

Cancel out b in both numerator and denominator.

$$\frac{9}{16b}\times \frac{32b^{2}h}{-27h^{2}b^{5}}$$

Cancel out hb^{2} in both numerator and denominator.

$$\frac{9}{16b}\times \frac{32}{-27hb^{3}}$$

Multiply \frac{9}{16b} times \frac{32}{-27hb^{3}} by multiplying numerator times numerator and denominator times denominator.

$$\frac{9\times 32}{16b\left(-27\right)hb^{3}}$$

Cancel out 9\times 16 in both numerator and denominator.

$$\frac{2}{-3bhb^{3}}$$

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

$$\frac{2}{-3b^{4}h}$$

Differentiate w.r.t. b

$$\frac{8}{3hb^{5}}$$