Solve 1/x-1/x+1=

$$\frac { 1 } { x } - \frac { 1 } { x + 1 } =$$

Evaluate

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

Short Solution Steps

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

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

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

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

Combine like terms in x+1-x.

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

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

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

Differentiate w.r.t. x

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

Steps Using Chain Rule

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

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

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

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

Combine like terms in x+1-x.

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

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.

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