Solve 4/x^2-4+2/x+2+1/2-x

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

Evaluate

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

Short Solution Steps

Factor x^{2}-4.

$$\frac{4}{\left(x-2\right)\left(x+2\right)}+\frac{2}{x+2}+\frac{1}{2-x}$$

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

$$\frac{4}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{1}{2-x}$$

Since \frac{4}{\left(x-2\right)\left(x+2\right)} and \frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)} have the same denominator, add them by adding their numerators.

$$\frac{4+2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{1}{2-x}$$

Do the multiplications in 4+2\left(x-2\right).

$$\frac{4+2x-4}{\left(x-2\right)\left(x+2\right)}+\frac{1}{2-x}$$

Combine like terms in 4+2x-4.

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

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

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

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

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

Do the multiplications in 2x-\left(x+2\right).

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

Combine like terms in 2x-x-2.

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

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

Factor x^{2}-4.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{4}{\left(x-2\right)\left(x+2\right)}+\frac{2}{x+2}+\frac{1}{2-x})$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{4}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{1}{2-x})$$

Since \frac{4}{\left(x-2\right)\left(x+2\right)} and \frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{4+2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{1}{2-x})$$

Do the multiplications in 4+2\left(x-2\right).

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{4+2x-4}{\left(x-2\right)\left(x+2\right)}+\frac{1}{2-x})$$

Combine like terms in 4+2x-4.

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

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

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

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

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

Do the multiplications in 2x-\left(x+2\right).

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

Combine like terms in 2x-x-2.

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

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.

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