Solve 1/1-4x^2/4x^2-1):(2/1+6x^-1/1-2x)=

$$\frac { 1 } { 1 } - \frac { 4 x ^ { 2 } } { 4 x ^ { 2 } - 1 } ) : ( \frac { 2 } { 1 } + \frac { 6 x ^ { - 1 } } { 1 - 2 x } ) =$$

Evaluate

$$-\frac{1}{4x^{2}-1}$$

Short Solution Steps

Divide 1 by 1 to get 1.

$$1-\frac{4x^{2}}{4x^{2}-1}$$

Factor 4x^{2}-1.

$$1-\frac{4x^{2}}{\left(2x-1\right)\left(2x+1\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}.

$$\frac{\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}-\frac{4x^{2}}{\left(2x-1\right)\left(2x+1\right)}$$

Since \frac{\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)} and \frac{4x^{2}}{\left(2x-1\right)\left(2x+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(2x-1\right)\left(2x+1\right)-4x^{2}}{\left(2x-1\right)\left(2x+1\right)}$$

Do the multiplications in \left(2x-1\right)\left(2x+1\right)-4x^{2}.

$$\frac{4x^{2}+2x-2x-1-4x^{2}}{\left(2x-1\right)\left(2x+1\right)}$$

Combine like terms in 4x^{2}+2x-2x-1-4x^{2}.

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

Expand \left(2x-1\right)\left(2x+1\right).

$$\frac{-1}{4x^{2}-1}$$

Differentiate w.r.t. x

$$\frac{8x}{\left(4x^{2}-1\right)^{2}}$$