Solve frac{32x+sqrt{3}}{3sqrt{2}}

$$\frac { 32 x + \sqrt { 3 } } { 3 \sqrt { 2 } }$$

Evaluate

$$\frac{\sqrt{2}\left(32x+\sqrt{3}\right)}{6}$$

Solution Steps

Rationalize the denominator of \frac{32x+\sqrt{3}}{3\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{\left(32x+\sqrt{3}\right)\sqrt{2}}{3\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(32x+\sqrt{3}\right)\sqrt{2}}{3\times 2}$$

Multiply 3 and 2 to get 6.

$$\frac{\left(32x+\sqrt{3}\right)\sqrt{2}}{6}$$

Use the distributive property to multiply 32x+\sqrt{3} by \sqrt{2}.

$$\frac{32x\sqrt{2}+\sqrt{3}\sqrt{2}}{6}$$

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

$$\frac{32x\sqrt{2}+\sqrt{6}}{6}$$

Differentiate w.r.t. x

$$\frac{16 \sqrt{2}}{3} = 7.542472332656508$$