Solve 18/x^2-4-2/x-2+x/x+2

$$1 \frac { 8 } { x ^ { 2 } - 4 } - \frac { 2 } { x - 2 } + \frac { x } { x + 2 }$$

Evaluate

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

Short Solution Steps

Express 1\times \frac{8}{x^{2}-4} as a single fraction.

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

Factor x^{2}-4.

$$\frac{8}{\left(x-2\right)\left(x+2\right)}-\frac{2}{x-2}+\frac{x}{x+2}$$

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

$$\frac{8}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x}{x+2}$$

Since \frac{8}{\left(x-2\right)\left(x+2\right)} and \frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{8-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x}{x+2}$$

Do the multiplications in 8-2\left(x+2\right).

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

Combine like terms in 8-2x-4.

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

Factor the expressions that are not already factored in \frac{4-2x}{\left(x-2\right)\left(x+2\right)}.

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

Extract the negative sign in 2-x.

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

Cancel out x-2 in both numerator and denominator.

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

Since \frac{-2}{x+2} and \frac{x}{x+2} have the same denominator, add them by adding their numerators.

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

Differentiate w.r.t. x

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

Steps Using Derivative Rule for Quotient

Express 1\times \frac{8}{x^{2}-4} as a single fraction.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8}{x^{2}-4}-\frac{2}{x-2}+\frac{x}{x+2})$$

Factor x^{2}-4.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8}{\left(x-2\right)\left(x+2\right)}-\frac{2}{x-2}+\frac{x}{x+2})$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x}{x+2})$$

Since \frac{8}{\left(x-2\right)\left(x+2\right)} and \frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x}{x+2})$$

Do the multiplications in 8-2\left(x+2\right).

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{8-2x-4}{\left(x-2\right)\left(x+2\right)}+\frac{x}{x+2})$$

Combine like terms in 8-2x-4.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{4-2x}{\left(x-2\right)\left(x+2\right)}+\frac{x}{x+2})$$

Factor the expressions that are not already factored in \frac{4-2x}{\left(x-2\right)\left(x+2\right)}.

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

Extract the negative sign in 2-x.

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

Cancel out x-2 in both numerator and denominator.

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

Since \frac{-2}{x+2} and \frac{x}{x+2} have the same denominator, add them by adding their numerators.

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

For any two differentiable functions, the derivative of the quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the denominator squared.

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

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}.

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

Do the arithmetic.

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

Expand using distributive property.

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

To multiply powers of the same base, add their exponents.

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

Remove unnecessary parentheses.

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

Combine like terms.

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

Subtract 1 from 1 and -2 from 2.

$$\frac{4x^{0}}{\left(x^{1}+2\right)^{2}}$$

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

$$\frac{4x^{0}}{\left(x+2\right)^{2}}$$

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

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

For any term t, t\times 1=t and 1t=t.

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