Solve 1/n+1-1/n+2

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

Evaluate

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

Short Solution Steps

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

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

Since \frac{n+2}{\left(n+1\right)\left(n+2\right)} and \frac{n+1}{\left(n+1\right)\left(n+2\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in n+2-\left(n+1\right).

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

Combine like terms in n+2-n-1.

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

Expand \left(n+1\right)\left(n+2\right).

$$\frac{1}{n^{2}+3n+2}$$

Differentiate w.r.t. n

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

Steps Using Chain Rule

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

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

Since \frac{n+2}{\left(n+1\right)\left(n+2\right)} and \frac{n+1}{\left(n+1\right)\left(n+2\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in n+2-\left(n+1\right).

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

Combine like terms in n+2-n-1.

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

Apply the distributive property by multiplying each term of n+1 by each term of n+2.

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

Combine 2n and n to get 3n.

$$\frac{\mathrm{d}}{\mathrm{d}n}(\frac{1}{n^{2}+3n+2})$$

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).

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

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}.

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

Simplify.

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

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

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

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

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