Solve -7/x^2-1-8/1-x^2

$$\frac { - 7 } { x ^ { 2 } - 1 } - \frac { 8 } { 1 - x ^ { 2 } }$$

Evaluate

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

Short Solution Steps

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

$$\frac{-7}{\left(x-1\right)\left(x+1\right)}-\frac{8}{\left(x-1\right)\left(-x-1\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-1\right)\left(x+1\right) and \left(x-1\right)\left(-x-1\right) is \left(x-1\right)\left(x+1\right). Multiply \frac{8}{\left(x-1\right)\left(-x-1\right)} times \frac{-1}{-1}.

$$\frac{-7}{\left(x-1\right)\left(x+1\right)}-\frac{8\left(-1\right)}{\left(x-1\right)\left(x+1\right)}$$

Since \frac{-7}{\left(x-1\right)\left(x+1\right)} and \frac{8\left(-1\right)}{\left(x-1\right)\left(x+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-7-8\left(-1\right)}{\left(x-1\right)\left(x+1\right)}$$

Do the multiplications in -7-8\left(-1\right).

$$\frac{-7+8}{\left(x-1\right)\left(x+1\right)}$$

Do the calculations in -7+8.

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

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

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

Differentiate w.r.t. x

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

Steps Using Chain Rule

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{-7}{\left(x-1\right)\left(x+1\right)}-\frac{8}{\left(x-1\right)\left(-x-1\right)})$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x-1\right)\left(x+1\right) and \left(x-1\right)\left(-x-1\right) is \left(x-1\right)\left(x+1\right). Multiply \frac{8}{\left(x-1\right)\left(-x-1\right)} times \frac{-1}{-1}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{-7}{\left(x-1\right)\left(x+1\right)}-\frac{8\left(-1\right)}{\left(x-1\right)\left(x+1\right)})$$

Since \frac{-7}{\left(x-1\right)\left(x+1\right)} and \frac{8\left(-1\right)}{\left(x-1\right)\left(x+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{-7-8\left(-1\right)}{\left(x-1\right)\left(x+1\right)})$$

Do the multiplications in -7-8\left(-1\right).

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{-7+8}{\left(x-1\right)\left(x+1\right)})$$

Do the calculations in -7+8.

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

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

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).

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

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}.

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

Simplify.

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

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

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