Solve 1/a^2+3ab-2/a^2-9b^2

$$\frac { 1 } { a ^ { 2 } + 3 a b } - \frac { 2 } { a ^ { 2 } - 9 b ^ { 2 } }$$

Evaluate

$$-\frac{1}{a\left(a-3b\right)}$$

Short Solution Steps

Factor a^{2}+3ab. Factor a^{2}-9b^{2}.

$$\frac{1}{a\left(a+3b\right)}-\frac{2}{\left(a-3b\right)\left(a+3b\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of a\left(a+3b\right) and \left(a-3b\right)\left(a+3b\right) is a\left(a-3b\right)\left(a+3b\right). Multiply \frac{1}{a\left(a+3b\right)} times \frac{a-3b}{a-3b}. Multiply \frac{2}{\left(a-3b\right)\left(a+3b\right)} times \frac{a}{a}.

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

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

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

Combine like terms in a-3b-2a.

$$\frac{-a-3b}{a\left(a-3b\right)\left(a+3b\right)}$$

Extract the negative sign in -a-3b.

$$\frac{-\left(a+3b\right)}{a\left(a-3b\right)\left(a+3b\right)}$$

Cancel out a+3b in both numerator and denominator.

$$\frac{-1}{a\left(a-3b\right)}$$

Expand a\left(a-3b\right).

$$\frac{-1}{a^{2}-3ab}$$

Differentiate w.r.t. a

$$\frac{2a-3b}{\left(a\left(a-3b\right)\right)^{2}}$$