Solve n/n+2/frac{n{n+2}+n}=

$$\frac { \frac { n } { n + 2 } } { \frac { n } { n + 2 } + n } =$$

Evaluate

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

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply n times \frac{n+2}{n+2}.

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

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

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

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

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

Combine like terms in n+n^{2}+2n.

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

Divide \frac{n}{n+2} by \frac{3n+n^{2}}{n+2} by multiplying \frac{n}{n+2} by the reciprocal of \frac{3n+n^{2}}{n+2}.

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

Cancel out n+2 in both numerator and denominator.

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

Factor the expressions that are not already factored.

$$\frac{n}{n\left(n+3\right)}$$

Cancel out n in both numerator and denominator.

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

Differentiate w.r.t. n

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

Steps Using Chain Rule

To add or subtract expressions, expand them to make their denominators the same. Multiply n times \frac{n+2}{n+2}.

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

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

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

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

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

Combine like terms in n+n^{2}+2n.

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

Divide \frac{n}{n+2} by \frac{3n+n^{2}}{n+2} by multiplying \frac{n}{n+2} by the reciprocal of \frac{3n+n^{2}}{n+2}.

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

Cancel out n+2 in both numerator and denominator.

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

Factor the expressions that are not already factored in \frac{n}{n^{2}+3n}.

$$\frac{\mathrm{d}}{\mathrm{d}n}(\frac{n}{n\left(n+3\right)})$$

Cancel out n in both numerator and denominator.

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

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).

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

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}.

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

Simplify.

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

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

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

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

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