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

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

Evaluate

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

Short Solution Steps

Factor x^{2}-1. Factor x^{2}+x.

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

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

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

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

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

Do the multiplications in 2x-\left(x-1\right).

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

Combine like terms in 2x-x+1.

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

Cancel out x+1 in both numerator and denominator.

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

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

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

Differentiate w.r.t. x

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

Steps Using Chain Rule

Factor x^{2}-1. Factor x^{2}+x.

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

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

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

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

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

Do the multiplications in 2x-\left(x-1\right).

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

Combine like terms in 2x-x+1.

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

Cancel out x+1 in both numerator and denominator.

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

Use the distributive property to multiply x by x-1.

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

If F is the composition of two differentiable functions f\left(u\right) and u=g\left(x\right), that is, if F\left(x\right)=f\left(g\left(x\right)\right), then the derivative of F is the derivative of f with respect to u times the derivative of g with respect to x, that is, \frac{\mathrm{d}}{\mathrm{d}x}(F)\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}(f)\left(g\left(x\right)\right)\frac{\mathrm{d}}{\mathrm{d}x}(g)\left(x\right).

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

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

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

Simplify.

$$\left(x^{2}-x^{1}\right)^{-2}\left(-2x^{1}+x^{0}\right)$$

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

$$\left(x^{2}-x\right)^{-2}\left(-2x+x^{0}\right)$$

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

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