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

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

Evaluate

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

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x-2}{x-2}.

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

Since \frac{x-2}{x-2} and \frac{2}{x-2} have the same denominator, add them by adding their numerators.

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

Combine like terms in x-2+2.

$$\frac{\frac{x}{x^{2}-4}}{\frac{x}{x-2}}$$

Divide \frac{x}{x^{2}-4} by \frac{x}{x-2} by multiplying \frac{x}{x^{2}-4} by the reciprocal of \frac{x}{x-2}.

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

Cancel out x in both numerator and denominator.

$$\frac{x-2}{x^{2}-4}$$

Factor the expressions that are not already factored.

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

Cancel out x-2 in both numerator and denominator.

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

Differentiate w.r.t. x

$$-\frac{1}{\left(x+2\right)^{2}}$$

Steps Using Chain Rule

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x-2}{x-2}.

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

Since \frac{x-2}{x-2} and \frac{2}{x-2} have the same denominator, add them by adding their numerators.

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

Combine like terms in x-2+2.

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

Divide \frac{x}{x^{2}-4} by \frac{x}{x-2} by multiplying \frac{x}{x^{2}-4} by the reciprocal of \frac{x}{x-2}.

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

Cancel out x in both numerator and denominator.

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

Factor the expressions that are not already factored in \frac{x-2}{x^{2}-4}.

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

Cancel out x-2 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{x+2})$$

If F is the composition of two differentiable functions f\left(u\right) and u=g\left(x\right), that is, if F\left(x\right)=f\left(g\left(x\right)\right), then the derivative of F is the derivative of f with respect to u times the derivative of g with respect to x, that is, \frac{\mathrm{d}}{\mathrm{d}x}(F)\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}(f)\left(g\left(x\right)\right)\frac{\mathrm{d}}{\mathrm{d}x}(g)\left(x\right).

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

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}.

$$-\left(x^{1}+2\right)^{-2}x^{1-1}$$

Simplify.

$$-x^{0}\left(x^{1}+2\right)^{-2}$$

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

$$-x^{0}\left(x+2\right)^{-2}$$

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

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