Solve (x/x+1-x/x-1)1-x^2/2

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

Evaluate

$$x$$

Short Solution Steps

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

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

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

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

Do the multiplications in x\left(x-1\right)-x\left(x+1\right).

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

Combine like terms in x^{2}-x-x^{2}-x.

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

Multiply \frac{-2x}{\left(x-1\right)\left(x+1\right)} times \frac{1-x^{2}}{2} by multiplying numerator times numerator and denominator times denominator.

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

Cancel out 2 in both numerator and denominator.

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

Factor the expressions that are not already factored.

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

Extract the negative sign in -1-x.

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

Cancel out \left(x-1\right)\left(x+1\right) in both numerator and denominator.

$$-\left(-1\right)x$$

Expand the expression.

$$x$$

Differentiate w.r.t. x

$$1$$

Steps Using Definition of a Derivative

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(\frac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\right)\times \frac{1-x^{2}}{2})$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x\left(x-1\right)-x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\times \frac{1-x^{2}}{2})$$

Do the multiplications in x\left(x-1\right)-x\left(x+1\right).

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x^{2}-x-x^{2}-x}{\left(x-1\right)\left(x+1\right)}\times \frac{1-x^{2}}{2})$$

Combine like terms in x^{2}-x-x^{2}-x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{-2x}{\left(x-1\right)\left(x+1\right)}\times \frac{1-x^{2}}{2})$$

Multiply \frac{-2x}{\left(x-1\right)\left(x+1\right)} times \frac{1-x^{2}}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{-2x\left(1-x^{2}\right)}{\left(x-1\right)\left(x+1\right)\times 2})$$

Cancel out 2 in both numerator and denominator.

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

Factor the expressions that are not already factored in \frac{-x\left(-x^{2}+1\right)}{\left(x-1\right)\left(x+1\right)}.

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

Extract the negative sign in -1-x.

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

Cancel out \left(x-1\right)\left(x+1\right) in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-\left(-1\right)x)$$

Expand the expression.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x)$$

The derivative of ax^{n} is nax^{n-1}.

$$x^{1-1}$$

Subtract 1 from 1.

$$x^{0}$$

For any term t except 0, t^{0}=1.

$$1$$