Solve 1/x+2+5/x-2

$$\frac { 1 } { x + 2 } + \frac { 5 } { x - 2 }$$

Evaluate

$$\frac{2\left(3x+4\right)}{x^{2}-4}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+2 and x-2 is \left(x-2\right)\left(x+2\right). Multiply \frac{1}{x+2} times \frac{x-2}{x-2}. Multiply \frac{5}{x-2} times \frac{x+2}{x+2}.

$$\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}$$

Since \frac{x-2}{\left(x-2\right)\left(x+2\right)} and \frac{5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)} have the same denominator, add them by adding their numerators.

$$\frac{x-2+5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}$$

Do the multiplications in x-2+5\left(x+2\right).

$$\frac{x-2+5x+10}{\left(x-2\right)\left(x+2\right)}$$

Combine like terms in x-2+5x+10.

$$\frac{6x+8}{\left(x-2\right)\left(x+2\right)}$$

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

$$\frac{6x+8}{x^{2}-4}$$

Differentiate w.r.t. x

$$-\frac{2\left(3x^{2}+8x+12\right)}{\left(x^{2}-4\right)^{2}}$$

Steps Using Derivative Rule for Quotient

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+2 and x-2 is \left(x-2\right)\left(x+2\right). Multiply \frac{1}{x+2} times \frac{x-2}{x-2}. Multiply \frac{5}{x-2} times \frac{x+2}{x+2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)})$$

Since \frac{x-2}{\left(x-2\right)\left(x+2\right)} and \frac{5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x-2+5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)})$$

Do the multiplications in x-2+5\left(x+2\right).

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x-2+5x+10}{\left(x-2\right)\left(x+2\right)})$$

Combine like terms in x-2+5x+10.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{6x+8}{\left(x-2\right)\left(x+2\right)})$$

Consider \left(x-2\right)\left(x+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{6x+8}{x^{2}-2^{2}})$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{6x+8}{x^{2}-4})$$

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.

$$\frac{\left(x^{2}-4\right)\frac{\mathrm{d}}{\mathrm{d}x}(6x^{1}+8)-\left(6x^{1}+8\right)\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-4)}{\left(x^{2}-4\right)^{2}}$$

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}.

$$\frac{\left(x^{2}-4\right)\times 6x^{1-1}-\left(6x^{1}+8\right)\times 2x^{2-1}}{\left(x^{2}-4\right)^{2}}$$

Do the arithmetic.

$$\frac{\left(x^{2}-4\right)\times 6x^{0}-\left(6x^{1}+8\right)\times 2x^{1}}{\left(x^{2}-4\right)^{2}}$$

Expand using distributive property.

$$\frac{x^{2}\times 6x^{0}-4\times 6x^{0}-\left(6x^{1}\times 2x^{1}+8\times 2x^{1}\right)}{\left(x^{2}-4\right)^{2}}$$

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

$$\frac{6x^{2}-4\times 6x^{0}-\left(6\times 2x^{1+1}+8\times 2x^{1}\right)}{\left(x^{2}-4\right)^{2}}$$

Do the arithmetic.

$$\frac{6x^{2}-24x^{0}-\left(12x^{2}+16x^{1}\right)}{\left(x^{2}-4\right)^{2}}$$

Remove unnecessary parentheses.

$$\frac{6x^{2}-24x^{0}-12x^{2}-16x^{1}}{\left(x^{2}-4\right)^{2}}$$

Combine like terms.

$$\frac{\left(6-12\right)x^{2}-24x^{0}-16x^{1}}{\left(x^{2}-4\right)^{2}}$$

Subtract 12 from 6.

$$\frac{-6x^{2}-24x^{0}-16x^{1}}{\left(x^{2}-4\right)^{2}}$$

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

$$\frac{-6x^{2}-24x^{0}-16x}{\left(x^{2}-4\right)^{2}}$$

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

$$\frac{-6x^{2}-24-16x}{\left(x^{2}-4\right)^{2}}$$