Solve frac{3sqrt{2}}{3sqrt{2}-2V}

$$\frac { 3 \sqrt { 2 } } { 3 \sqrt { 2 } - 2 V }$$

Evaluate

$$\frac{3\sqrt{2}}{-2V+3\sqrt{2}}$$

Short Solution Steps

Rationalize the denominator of \frac{3\sqrt{2}}{3\sqrt{2}-2V} by multiplying numerator and denominator by 3\sqrt{2}+2V.

$$\frac{3\sqrt{2}\left(3\sqrt{2}+2V\right)}{\left(3\sqrt{2}-2V\right)\left(3\sqrt{2}+2V\right)}$$

Consider \left(3\sqrt{2}-2V\right)\left(3\sqrt{2}+2V\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{3\sqrt{2}\left(3\sqrt{2}+2V\right)}{\left(3\sqrt{2}\right)^{2}-\left(-2V\right)^{2}}$$

Expand \left(3\sqrt{2}\right)^{2}.

$$\frac{3\sqrt{2}\left(3\sqrt{2}+2V\right)}{3^{2}\left(\sqrt{2}\right)^{2}-\left(-2V\right)^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{3\sqrt{2}\left(3\sqrt{2}+2V\right)}{9\left(\sqrt{2}\right)^{2}-\left(-2V\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{3\sqrt{2}\left(3\sqrt{2}+2V\right)}{9\times 2-\left(-2V\right)^{2}}$$

Multiply 9 and 2 to get 18.

$$\frac{3\sqrt{2}\left(3\sqrt{2}+2V\right)}{18-\left(-2V\right)^{2}}$$

Expand \left(-2V\right)^{2}.

$$\frac{3\sqrt{2}\left(3\sqrt{2}+2V\right)}{18-\left(-2\right)^{2}V^{2}}$$

Calculate -2 to the power of 2 and get 4.

$$\frac{3\sqrt{2}\left(3\sqrt{2}+2V\right)}{18-4V^{2}}$$

Use the distributive property to multiply 3\sqrt{2} by 3\sqrt{2}+2V.

$$\frac{9\left(\sqrt{2}\right)^{2}+6V\sqrt{2}}{18-4V^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{9\times 2+6V\sqrt{2}}{18-4V^{2}}$$

Multiply 9 and 2 to get 18.

$$\frac{18+6V\sqrt{2}}{18-4V^{2}}$$

Factor the expressions that are not already factored.

$$\frac{6\left(\sqrt{2}V+3\right)}{2\left(-2V^{2}+9\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{3\left(\sqrt{2}V+3\right)}{-2V^{2}+9}$$

Expand the expression.

$$\frac{3\sqrt{2}V+9}{-2V^{2}+9}$$

Differentiate w.r.t. V

$$\frac{6\sqrt{2}}{\left(-2V+3\sqrt{2}\right)^{2}}$$