Solve =1/2xsqrt{2-sqrt{3}}

$$= \frac { 1 } { 2 x \sqrt { 2 } - \sqrt { 3 } }$$

Evaluate

$$\frac{1}{2\sqrt{2}x-\sqrt{3}}$$

Solution Steps

Rationalize the denominator of \frac{1}{2x\sqrt{2}-\sqrt{3}} by multiplying numerator and denominator by 2x\sqrt{2}+\sqrt{3}.

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{2} is 2.

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

Multiply 4 and 2 to get 8.

$$\frac{2x\sqrt{2}+\sqrt{3}}{8x^{2}-\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{2x\sqrt{2}+\sqrt{3}}{8x^{2}-3}$$

Differentiate w.r.t. x

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