Solve 2b/b^2-a^2div(1/a-b+1/a+b)

$$\frac { 2 b } { b ^ { 2 } - a ^ { 2 } } \div ( \frac { 1 } { a - b } + \frac { 1 } { a + b } )$$

Evaluate

$$-\frac{b}{a}$$

Short Solution Steps

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

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

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

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

Combine like terms in a+b+a-b.

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

Divide \frac{2b}{b^{2}-a^{2}} by \frac{2a}{\left(a+b\right)\left(a-b\right)} by multiplying \frac{2b}{b^{2}-a^{2}} by the reciprocal of \frac{2a}{\left(a+b\right)\left(a-b\right)}.

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

Cancel out 2 in both numerator and denominator.

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

Factor the expressions that are not already factored.

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

Extract the negative sign in a+b.

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

Cancel out \left(a-b\right)\left(-a-b\right) in both numerator and denominator.

$$\frac{-b}{a}$$

Differentiate w.r.t. b

$$-\frac{1}{a}$$