Solve 2x/x^2-4^2div(1/4-x+1/4+x)

$$\frac { 2 x } { x ^ { 2 } - 4 ^ { 2 } } \div ( \frac { 1 } { 4 - x } + \frac { 1 } { 4 + x } )$$

Evaluate

$$-\frac{x}{4}$$

Short Solution Steps

Calculate 4 to the power of 2 and get 16.

$$\frac{\frac{2x}{x^{2}-16}}{\frac{1}{4-x}+\frac{1}{4+x}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4-x and 4+x is \left(x+4\right)\left(-x+4\right). Multiply \frac{1}{4-x} times \frac{x+4}{x+4}. Multiply \frac{1}{4+x} times \frac{-x+4}{-x+4}.

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

Since \frac{x+4}{\left(x+4\right)\left(-x+4\right)} and \frac{-x+4}{\left(x+4\right)\left(-x+4\right)} have the same denominator, add them by adding their numerators.

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

Combine like terms in x+4-x+4.

$$\frac{\frac{2x}{x^{2}-16}}{\frac{8}{\left(x+4\right)\left(-x+4\right)}}$$

Divide \frac{2x}{x^{2}-16} by \frac{8}{\left(x+4\right)\left(-x+4\right)} by multiplying \frac{2x}{x^{2}-16} by the reciprocal of \frac{8}{\left(x+4\right)\left(-x+4\right)}.

$$\frac{2x\left(x+4\right)\left(-x+4\right)}{\left(x^{2}-16\right)\times 8}$$

Cancel out 2 in both numerator and denominator.

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

Factor the expressions that are not already factored.

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

Extract the negative sign in 4-x.

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

Cancel out \left(x-4\right)\left(x+4\right) in both numerator and denominator.

$$\frac{-x}{4}$$

Differentiate w.r.t. x

$$-\frac{1}{4} = -0.25$$