Solve (x-1/sqrt{2})^2-(x-1/sqrt{2})^2-(2x+sqrt{2})

$$( x - \frac { 1 } { \sqrt { 2 } } ) ^ { 2 } - ( x - \frac { 1 } { \sqrt { 2 } } ) ^ { 2 } - ( 2 x + \sqrt { 2 } )$$

Evaluate

$$-2x-\sqrt{2}$$

Short Solution Steps

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

$$\left(x-\frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}\right)^{2}-\left(x-\frac{1}{\sqrt{2}}\right)^{2}-\left(2x+\sqrt{2}\right)$$

The square of \sqrt{2} is 2.

$$\left(x-\frac{\sqrt{2}}{2}\right)^{2}-\left(x-\frac{1}{\sqrt{2}}\right)^{2}-\left(2x+\sqrt{2}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{2}{2}.

$$\left(\frac{2x}{2}-\frac{\sqrt{2}}{2}\right)^{2}-\left(x-\frac{1}{\sqrt{2}}\right)^{2}-\left(2x+\sqrt{2}\right)$$

Since \frac{2x}{2} and \frac{\sqrt{2}}{2} have the same denominator, subtract them by subtracting their numerators.

$$\left(\frac{2x-\sqrt{2}}{2}\right)^{2}-\left(x-\frac{1}{\sqrt{2}}\right)^{2}-\left(2x+\sqrt{2}\right)$$

To raise \frac{2x-\sqrt{2}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(2x-\sqrt{2}\right)^{2}}{2^{2}}-\left(x-\frac{1}{\sqrt{2}}\right)^{2}-\left(2x+\sqrt{2}\right)$$

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

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

The square of \sqrt{2} is 2.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{2}{2}.

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

Since \frac{2x}{2} and \frac{\sqrt{2}}{2} have the same denominator, subtract them by subtracting their numerators.

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

To raise \frac{2x-\sqrt{2}}{2} to a power, raise both numerator and denominator to the power and then divide.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-\sqrt{2}\right)^{2}.

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

The square of \sqrt{2} is 2.

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

Calculate 2 to the power of 2 and get 4.

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

To add or subtract expressions, expand them to make their denominators the same. Expand 2^{2}.

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

Since \frac{\left(2x-\sqrt{2}\right)^{2}}{4} and \frac{4x^{2}-4x\sqrt{2}+2}{4} have the same denominator, subtract them by subtracting their numerators.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-\sqrt{2}\right)^{2}.

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

The square of \sqrt{2} is 2.

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

To find the opposite of 2x+\sqrt{2}, find the opposite of each term.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply -2x-\sqrt{2} times \frac{4}{4}.

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

Since \frac{4x^{2}-4x\sqrt{2}+2-\left(4x^{2}-4x\sqrt{2}+2\right)}{4} and \frac{4\left(-2x-\sqrt{2}\right)}{4} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 4x^{2}-4x\sqrt{2}+2-\left(4x^{2}-4x\sqrt{2}+2\right)+4\left(-2x-\sqrt{2}\right).

$$\frac{4x^{2}-4x\sqrt{2}+2-4x^{2}+4x\sqrt{2}-2-8x-4\sqrt{2}}{4}$$

Combine like terms in 4x^{2}-4x\sqrt{2}+2-4x^{2}+4x\sqrt{2}-2-8x-4\sqrt{2}.

$$\frac{-4\sqrt{2}-8x}{4}$$

Divide each term of -4\sqrt{2}-8x by 4 to get -\sqrt{2}-2x.

$$-\sqrt{2}-2x$$

Differentiate w.r.t. x

$$-2$$