Solve 2/x^2-2x-3+1/1-x^2=

$$\frac { 2 } { x ^ { 2 } - 2 x - 3 } + \frac { 1 } { 1 - x ^ { 2 } } =$$

Evaluate

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

Short Solution Steps

Factor x^{2}-2x-3. Factor 1-x^{2}.

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

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

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

Since \frac{2\left(x-1\right)}{\left(x-3\right)\left(x-1\right)\left(x+1\right)} and \frac{-\left(x-3\right)}{\left(x-3\right)\left(x-1\right)\left(x+1\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 2\left(x-1\right)-\left(x-3\right).

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

Combine like terms in 2x-2-x+3.

$$\frac{x+1}{\left(x-3\right)\left(x-1\right)\left(x+1\right)}$$

Cancel out x+1 in both numerator and denominator.

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

Expand \left(x-3\right)\left(x-1\right).

$$\frac{1}{x^{2}-4x+3}$$

Differentiate w.r.t. x

$$\frac{2\left(2-x\right)}{\left(\left(1-x\right)\left(x-3\right)\right)^{2}}$$

Steps Using Chain Rule

Factor x^{2}-2x-3. Factor 1-x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2}{\left(x-3\right)\left(x+1\right)}+\frac{1}{\left(x-1\right)\left(-x-1\right)})$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2\left(x-1\right)}{\left(x-3\right)\left(x-1\right)\left(x+1\right)}+\frac{-\left(x-3\right)}{\left(x-3\right)\left(x-1\right)\left(x+1\right)})$$

Since \frac{2\left(x-1\right)}{\left(x-3\right)\left(x-1\right)\left(x+1\right)} and \frac{-\left(x-3\right)}{\left(x-3\right)\left(x-1\right)\left(x+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2\left(x-1\right)-\left(x-3\right)}{\left(x-3\right)\left(x-1\right)\left(x+1\right)})$$

Do the multiplications in 2\left(x-1\right)-\left(x-3\right).

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2x-2-x+3}{\left(x-3\right)\left(x-1\right)\left(x+1\right)})$$

Combine like terms in 2x-2-x+3.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x+1}{\left(x-3\right)\left(x-1\right)\left(x+1\right)})$$

Cancel out x+1 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{\left(x-3\right)\left(x-1\right)})$$

Use the distributive property to multiply x-3 by x-1 and combine like terms.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{x^{2}-4x+3})$$

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).

$$-\left(x^{2}-4x^{1}+3\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-4x^{1}+3)$$

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}.

$$-\left(x^{2}-4x^{1}+3\right)^{-2}\left(2x^{2-1}-4x^{1-1}\right)$$

Simplify.

$$\left(x^{2}-4x^{1}+3\right)^{-2}\left(-2x^{1}+4x^{0}\right)$$

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

$$\left(x^{2}-4x+3\right)^{-2}\left(-2x+4x^{0}\right)$$

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

$$\left(x^{2}-4x+3\right)^{-2}\left(-2x+4\times 1\right)$$

For any term t, t\times 1=t and 1t=t.

$$\left(x^{2}-4x+3\right)^{-2}\left(-2x+4\right)$$