Solve phi(2sqrt{48}-3sqrt{27})divsqrt{6}

$$\phi ( 2 \sqrt { 48 } - 3 \sqrt { 27 } ) \div \sqrt { 6 }$$

Evaluate

$$-\frac{\sqrt{2}ϕ}{2}$$

Solution Steps

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$\frac{ϕ\left(2\times 4\sqrt{3}-3\sqrt{27}\right)}{\sqrt{6}}$$

Multiply 2 and 4 to get 8.

$$\frac{ϕ\left(8\sqrt{3}-3\sqrt{27}\right)}{\sqrt{6}}$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$\frac{ϕ\left(8\sqrt{3}-3\times 3\sqrt{3}\right)}{\sqrt{6}}$$

Multiply -3 and 3 to get -9.

$$\frac{ϕ\left(8\sqrt{3}-9\sqrt{3}\right)}{\sqrt{6}}$$

Combine 8\sqrt{3} and -9\sqrt{3} to get -\sqrt{3}.

$$\frac{ϕ\left(-1\right)\sqrt{3}}{\sqrt{6}}$$

Rationalize the denominator of \frac{ϕ\left(-1\right)\sqrt{3}}{\sqrt{6}} by multiplying numerator and denominator by \sqrt{6}.

$$\frac{ϕ\left(-1\right)\sqrt{3}\sqrt{6}}{\left(\sqrt{6}\right)^{2}}$$

The square of \sqrt{6} is 6.

$$\frac{ϕ\left(-1\right)\sqrt{3}\sqrt{6}}{6}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$\frac{ϕ\left(-1\right)\sqrt{3}\sqrt{3}\sqrt{2}}{6}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$\frac{ϕ\left(-1\right)\times 3\sqrt{2}}{6}$$

Multiply -1 and 3 to get -3.

$$\frac{ϕ\left(-3\right)\sqrt{2}}{6}$$

Divide ϕ\left(-3\right)\sqrt{2} by 6 to get ϕ\left(-\frac{1}{2}\right)\sqrt{2}.

$$ϕ\left(-\frac{1}{2}\right)\sqrt{2}$$

Differentiate w.r.t. ϕ

$$-\frac{\sqrt{2}}{2} = -0.7071067811865476$$