Solve 4x^3/20y^2*8y^2/9x^2=

$$\frac { 4 x ^ { 3 } } { 20 y ^ { 2 } } \times \frac { 8 y ^ { 2 } } { 9 x ^ { 2 } } =$$

Evaluate

$$\frac{8x}{45}$$

Solution Steps

Cancel out 4 in both numerator and denominator.

$$\frac{x^{3}}{5y^{2}}\times \frac{8y^{2}}{9x^{2}}$$

Multiply \frac{x^{3}}{5y^{2}} times \frac{8y^{2}}{9x^{2}} by multiplying numerator times numerator and denominator times denominator.

$$\frac{x^{3}\times 8y^{2}}{5y^{2}\times 9x^{2}}$$

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

$$\frac{8x}{5\times 9}$$

Multiply 5 and 9 to get 45.

$$\frac{8x}{45}$$

Differentiate w.r.t. x

$$\frac{8}{45} = 0.17777777777777778$$