Solve 1/x+3+1/x-3-6/x^2-9

$$\frac { 1 } { x + 3 } + \frac { 1 } { x - 3 } - \frac { 6 } { x ^ { 2 } - 9 }$$

Evaluate

$$\frac{2}{x+3}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+3 and x-3 is \left(x-3\right)\left(x+3\right). Multiply \frac{1}{x+3} times \frac{x-3}{x-3}. Multiply \frac{1}{x-3} times \frac{x+3}{x+3}.

$$\frac{x-3}{\left(x-3\right)\left(x+3\right)}+\frac{x+3}{\left(x-3\right)\left(x+3\right)}-\frac{6}{x^{2}-9}$$

Since \frac{x-3}{\left(x-3\right)\left(x+3\right)} and \frac{x+3}{\left(x-3\right)\left(x+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{x-3+x+3}{\left(x-3\right)\left(x+3\right)}-\frac{6}{x^{2}-9}$$

Combine like terms in x-3+x+3.

$$\frac{2x}{\left(x-3\right)\left(x+3\right)}-\frac{6}{x^{2}-9}$$

Factor x^{2}-9.

$$\frac{2x}{\left(x-3\right)\left(x+3\right)}-\frac{6}{\left(x-3\right)\left(x+3\right)}$$

Since \frac{2x}{\left(x-3\right)\left(x+3\right)} and \frac{6}{\left(x-3\right)\left(x+3\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2x-6}{\left(x-3\right)\left(x+3\right)}$$

Factor the expressions that are not already factored in \frac{2x-6}{\left(x-3\right)\left(x+3\right)}.

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

Cancel out x-3 in both numerator and denominator.

$$\frac{2}{x+3}$$

Differentiate w.r.t. x

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