Solve sqrt{1/3}xsqrt{12}+sqrt{6}divsqrt{2}*sqrt{2}

$$\sqrt { \frac { 1 } { 3 } } x \sqrt { 12 } + \sqrt { 6 } \div \sqrt { 2 } \cdot \sqrt { 2 }$$

Evaluate

$$2x+\sqrt{6}$$

Short Solution Steps

Rewrite the square root of the division \sqrt{\frac{1}{3}} as the division of square roots \frac{\sqrt{1}}{\sqrt{3}}.

$$\frac{\sqrt{1}}{\sqrt{3}}x\sqrt{12}+\frac{\sqrt{6}}{\sqrt{2}}\sqrt{2}$$

Calculate the square root of 1 and get 1.

$$\frac{1}{\sqrt{3}}x\sqrt{12}+\frac{\sqrt{6}}{\sqrt{2}}\sqrt{2}$$

Rationalize the denominator of \frac{1}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}x\sqrt{12}+\frac{\sqrt{6}}{\sqrt{2}}\sqrt{2}$$

The square of \sqrt{3} is 3.

$$\frac{\sqrt{3}}{3}x\sqrt{12}+\frac{\sqrt{6}}{\sqrt{2}}\sqrt{2}$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\frac{\sqrt{3}}{3}x\times 2\sqrt{3}+\frac{\sqrt{6}}{\sqrt{2}}\sqrt{2}$$

Express \frac{\sqrt{3}}{3}x as a single fraction.

$$\frac{\sqrt{3}x}{3}\times 2\sqrt{3}+\frac{\sqrt{6}}{\sqrt{2}}\sqrt{2}$$

Express \frac{\sqrt{3}x}{3}\times 2 as a single fraction.

$$\frac{\sqrt{3}x\times 2}{3}\sqrt{3}+\frac{\sqrt{6}}{\sqrt{2}}\sqrt{2}$$

Express \frac{\sqrt{3}x\times 2}{3}\sqrt{3} as a single fraction.

$$\frac{\sqrt{3}x\times 2\sqrt{3}}{3}+\frac{\sqrt{6}}{\sqrt{2}}\sqrt{2}$$

Rewrite the division of square roots \frac{\sqrt{6}}{\sqrt{2}} as the square root of the division \sqrt{\frac{6}{2}} and perform the division.

$$\frac{\sqrt{3}x\times 2\sqrt{3}}{3}+\sqrt{3}\sqrt{2}$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$\frac{\sqrt{3}x\times 2\sqrt{3}}{3}+\sqrt{6}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{6} times \frac{3}{3}.

$$\frac{\sqrt{3}x\times 2\sqrt{3}}{3}+\frac{3\sqrt{6}}{3}$$

Since \frac{\sqrt{3}x\times 2\sqrt{3}}{3} and \frac{3\sqrt{6}}{3} have the same denominator, add them by adding their numerators.

$$\frac{\sqrt{3}x\times 2\sqrt{3}+3\sqrt{6}}{3}$$

Do the multiplications in \sqrt{3}x\times 2\sqrt{3}+3\sqrt{6}.

$$\frac{6x+3\sqrt{6}}{3}$$

Divide each term of 6x+3\sqrt{6} by 3 to get 2x+\sqrt{6}.

$$2x+\sqrt{6}$$

Differentiate w.r.t. x

$$2$$