Solve 1/x+1/x-1

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

Evaluate

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

Solution Steps

Express \frac{\frac{1}{x+1}}{x-1} as a single fraction.

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

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}.

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

Calculate 1 to the power of 2 and get 1.

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

Differentiate w.r.t. x

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

Steps Using Chain Rule

Express \frac{\frac{1}{x+1}}{x-1} as a single fraction.

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

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}.

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

Calculate 1 to the power of 2 and get 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.

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