Solve 1/n-1/n+1

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

Evaluate

$$\frac{1}{n\left(n+1\right)}$$

Short Solution Steps

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

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

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

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

Combine like terms in n+1-n.

$$\frac{1}{n\left(n+1\right)}$$

Expand n\left(n+1\right).

$$\frac{1}{n^{2}+n}$$

Differentiate w.r.t. n

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

Steps Using Chain Rule

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

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

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

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

Combine like terms in n+1-n.

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

Use the distributive property to multiply n by n+1.

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

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).

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

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}.

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

Simplify.

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

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

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

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

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