Find the Derivative - d/dn (1+1/n)^n

${(1 + \frac{1}{n})}^{n}$

Step 1

Use the properties of logarithms to simplify the differentiation.

Rewrite ${(1 + \frac{1}{n})}^{n}$ as $e^{\ln ({(1 + \frac{1}{n})}^{n})}$ .

$\frac{d}{d n} [e^{\ln ({(1 + \frac{1}{n})}^{n})}]$

Expand $\ln ({(1 + \frac{1}{n})}^{n})$ by moving $n$ outside the logarithm.

$\frac{d}{d n} [e^{n \ln (1 + \frac{1}{n})}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d n} [f (g (n))]$ is $f' (g (n)) g' (n)$ where $f (n) = e^{n}$ and $g (n) = n \ln (1 + \frac{1}{n})$ .

To apply the Chain Rule, set $u_{1}$ as $n \ln (1 + \frac{1}{n})$ .

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d n} [n \ln (1 + \frac{1}{n})]$

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$e^{u_{1}} \frac{d}{d n} [n \ln (1 + \frac{1}{n})]$

Replace all occurrences of $u_{1}$ with $n \ln (1 + \frac{1}{n})$ .

$e^{n \ln (1 + \frac{1}{n})} \frac{d}{d n} [n \ln (1 + \frac{1}{n})]$

Step 3

Differentiate using the Product Rule which states that $\frac{d}{d n} [f (n) g (n)]$ is $f (n) \frac{d}{d n} [g (n)] + g (n) \frac{d}{d n} [f (n)]$ where $f (n) = n$ and $g (n) = \ln (1 + \frac{1}{n})$ .

$e^{n \ln (1 + \frac{1}{n})} (n \frac{d}{d n} [\ln (1 + \frac{1}{n})] + \ln (1 + \frac{1}{n}) \frac{d}{d n} [n])$

Step 4

Differentiate using the chain rule, which states that $\frac{d}{d n} [f (g (n))]$ is $f' (g (n)) g' (n)$ where $f (n) = \ln (n)$ and $g (n) = 1 + \frac{1}{n}$ .

To apply the Chain Rule, set $u_{2}$ as $1 + \frac{1}{n}$ .

$e^{n \ln (1 + \frac{1}{n})} (n (\frac{d}{d u_{2}} [\ln (u_{2})] \frac{d}{d n} [1 + \frac{1}{n}]) + \ln (1 + \frac{1}{n}) \frac{d}{d n} [n])$

The derivative of $\ln (u_{2})$ with respect to $u_{2}$ is $\frac{1}{u_{2}}$ .

$e^{n \ln (1 + \frac{1}{n})} (n (\frac{1}{u_{2}} \frac{d}{d n} [1 + \frac{1}{n}]) + \ln (1 + \frac{1}{n}) \frac{d}{d n} [n])$

Replace all occurrences of $u_{2}$ with $1 + \frac{1}{n}$ .

$e^{n \ln (1 + \frac{1}{n})} (n (\frac{1}{1 + \frac{1}{n}} \frac{d}{d n} [1 + \frac{1}{n}]) + \ln (1 + \frac{1}{n}) \frac{d}{d n} [n])$

Step 5

Differentiate.

Combine $\frac{1}{1 + \frac{1}{n}}$ and $n$ .

$e^{n \ln (1 + \frac{1}{n})} (\frac{n}{1 + \frac{1}{n}} \frac{d}{d n} [1 + \frac{1}{n}] + \ln (1 + \frac{1}{n}) \frac{d}{d n} [n])$

By the Sum Rule, the derivative of $1 + \frac{1}{n}$ with respect to $n$ is $\frac{d}{d n} [1] + \frac{d}{d n} [\frac{1}{n}]$ .

$e^{n \ln (1 + \frac{1}{n})} (\frac{n}{1 + \frac{1}{n}} (\frac{d}{d n} [1] + \frac{d}{d n} [\frac{1}{n}]) + \ln (1 + \frac{1}{n}) \frac{d}{d n} [n])$

Since $1$ is constant with respect to $n$ , the derivative of $1$ with respect to $n$ is $0$ .

$e^{n \ln (1 + \frac{1}{n})} (\frac{n}{1 + \frac{1}{n}} (0 + \frac{d}{d n} [\frac{1}{n}]) + \ln (1 + \frac{1}{n}) \frac{d}{d n} [n])$

Simplify the expression.

Add $0$ and $\frac{d}{d n} [\frac{1}{n}]$ .

$e^{n \ln (1 + \frac{1}{n})} (\frac{n}{1 + \frac{1}{n}} \frac{d}{d n} [\frac{1}{n}] + \ln (1 + \frac{1}{n}) \frac{d}{d n} [n])$

Rewrite $\frac{1}{n}$ as $n^{- 1}$ .

$e^{n \ln (1 + \frac{1}{n})} (\frac{n}{1 + \frac{1}{n}} \frac{d}{d n} [n^{- 1}] + \ln (1 + \frac{1}{n}) \frac{d}{d n} [n])$

Differentiate using the Power Rule which states that $\frac{d}{d n} [n^{n}]$ is $n \cdot n^{n - 1}$ where $n = - 1$ .

$e^{n \ln (1 + \frac{1}{n})} (\frac{n}{1 + \frac{1}{n}} (- n^{- 2}) + \ln (1 + \frac{1}{n}) \frac{d}{d n} [n])$

Combine $\frac{n}{1 + \frac{1}{n}}$ and $n^{- 2}$ .

$e^{n \ln (1 + \frac{1}{n})} (- \frac{n \cdot n^{- 2}}{1 + \frac{1}{n}} + \ln (1 + \frac{1}{n}) \frac{d}{d n} [n])$

Step 6

Multiply $n$ by $n^{- 2}$ by adding the exponents.

Multiply $n$ by $n^{- 2}$ .

Raise $n$ to the power of $1$ .

$e^{n \ln (1 + \frac{1}{n})} (- \frac{n^{1} n^{- 2}}{1 + \frac{1}{n}} + \ln (1 + \frac{1}{n}) \frac{d}{d n} [n])$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$e^{n \ln (1 + \frac{1}{n})} (- \frac{n^{1 - 2}}{1 + \frac{1}{n}} + \ln (1 + \frac{1}{n}) \frac{d}{d n} [n])$

Subtract $2$ from $1$ .

$e^{n \ln (1 + \frac{1}{n})} (- \frac{n^{- 1}}{1 + \frac{1}{n}} + \ln (1 + \frac{1}{n}) \frac{d}{d n} [n])$

Step 7

Move $n^{- 1}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$e^{n \ln (1 + \frac{1}{n})} (- \frac{1}{(1 + \frac{1}{n}) n} + \ln (1 + \frac{1}{n}) \frac{d}{d n} [n])$

Step 8

Differentiate using the Power Rule which states that $\frac{d}{d n} [n^{n}]$ is $n \cdot n^{n - 1}$ where $n = 1$ .

$e^{n \ln (1 + \frac{1}{n})} (- \frac{1}{(1 + \frac{1}{n}) n} + \ln (1 + \frac{1}{n}) \cdot 1)$

Step 9

Multiply $\ln (1 + \frac{1}{n})$ by $1$ .

$e^{n \ln (1 + \frac{1}{n})} (- \frac{1}{(1 + \frac{1}{n}) n} + \ln (1 + \frac{1}{n}))$

Step 10

To write $\ln (1 + \frac{1}{n})$ as a fraction with a common denominator, multiply by $\frac{(1 + \frac{1}{n}) n}{(1 + \frac{1}{n}) n}$ .

$e^{n \ln (1 + \frac{1}{n})} (- \frac{1}{(1 + \frac{1}{n}) n} + \ln (1 + \frac{1}{n}) \cdot \frac{(1 + \frac{1}{n}) n}{(1 + \frac{1}{n}) n})$

Step 11

Combine $\ln (1 + \frac{1}{n})$ and $\frac{(1 + \frac{1}{n}) n}{(1 + \frac{1}{n}) n}$ .

$e^{n \ln (1 + \frac{1}{n})} (- \frac{1}{(1 + \frac{1}{n}) n} + \frac{\ln (1 + \frac{1}{n}) ((1 + \frac{1}{n}) n)}{(1 + \frac{1}{n}) n})$

Step 12

Combine the numerators over the common denominator.

$e^{n \ln (1 + \frac{1}{n})} \frac{- 1 + \ln (1 + \frac{1}{n}) ((1 + \frac{1}{n}) n)}{(1 + \frac{1}{n}) n}$

Step 13

Combine $e^{n \ln (1 + \frac{1}{n})}$ and $\frac{- 1 + \ln (1 + \frac{1}{n}) ((1 + \frac{1}{n}) n)}{(1 + \frac{1}{n}) n}$ .

$\frac{e^{n \ln (1 + \frac{1}{n})} (- 1 + \ln (1 + \frac{1}{n}) ((1 + \frac{1}{n}) n))}{(1 + \frac{1}{n}) n}$

Step 14

Simplify.

Apply the distributive property.

$\frac{e^{n \ln (1 + \frac{1}{n})} (- 1 + \ln (1 + \frac{1}{n}) (1 n + \frac{1}{n} n))}{(1 + \frac{1}{n}) n}$

Apply the distributive property.

$\frac{e^{n \ln (1 + \frac{1}{n})} (- 1 + \ln (1 + \frac{1}{n}) (1 n + \frac{1}{n} n))}{1 n + \frac{1}{n} n}$

Simplify the numerator.

Simplify each term.

Multiply $n$ by $1$ .

$\frac{e^{n \ln (1 + \frac{1}{n})} (- 1 + \ln (1 + \frac{1}{n}) (n + \frac{1}{n} n))}{1 n + \frac{1}{n} n}$

Cancel the common factor of $n$ .

Cancel the common factor.

$\frac{e^{n \ln (1 + \frac{1}{n})} (- 1 + \ln (1 + \frac{1}{n}) (n + \frac{1}{n} n))}{1 n + \frac{1}{n} n}$

Rewrite the expression.

$\frac{e^{n \ln (1 + \frac{1}{n})} (- 1 + \ln (1 + \frac{1}{n}) (n + 1))}{1 n + \frac{1}{n} n}$

Apply the distributive property.

$\frac{e^{n \ln (1 + \frac{1}{n})} (- 1 + \ln (1 + \frac{1}{n}) n + \ln (1 + \frac{1}{n}) \cdot 1)}{1 n + \frac{1}{n} n}$

Multiply $\ln (1 + \frac{1}{n})$ by $1$ .

$\frac{e^{n \ln (1 + \frac{1}{n})} (- 1 + \ln (1 + \frac{1}{n}) n + \ln (1 + \frac{1}{n}))}{1 n + \frac{1}{n} n}$

Apply the distributive property.

$\frac{e^{n \ln (1 + \frac{1}{n})} \cdot - 1 + e^{n \ln (1 + \frac{1}{n})} (\ln (1 + \frac{1}{n}) n) + e^{n \ln (1 + \frac{1}{n})} \ln (1 + \frac{1}{n})}{1 n + \frac{1}{n} n}$

Move $- 1$ to the left of $e^{n \ln (1 + \frac{1}{n})}$ .

$\frac{- 1 \cdot e^{n \ln (1 + \frac{1}{n})} + e^{n \ln (1 + \frac{1}{n})} (\ln (1 + \frac{1}{n}) n) + e^{n \ln (1 + \frac{1}{n})} \ln (1 + \frac{1}{n})}{1 n + \frac{1}{n} n}$

Rewrite $- 1 e^{n \ln (1 + \frac{1}{n})}$ as $- e^{n \ln (1 + \frac{1}{n})}$ .

$\frac{- e^{n \ln (1 + \frac{1}{n})} + e^{n \ln (1 + \frac{1}{n})} (\ln (1 + \frac{1}{n}) n) + e^{n \ln (1 + \frac{1}{n})} \ln (1 + \frac{1}{n})}{1 n + \frac{1}{n} n}$

Reorder factors in $- e^{n \ln (1 + \frac{1}{n})} + e^{n \ln (1 + \frac{1}{n})} (\ln (1 + \frac{1}{n}) n) + e^{n \ln (1 + \frac{1}{n})} \ln (1 + \frac{1}{n})$ .

$\frac{- e^{n \ln (1 + \frac{1}{n})} + n e^{n \ln (1 + \frac{1}{n})} \ln (1 + \frac{1}{n}) + e^{n \ln (1 + \frac{1}{n})} \ln (1 + \frac{1}{n})}{1 n + \frac{1}{n} n}$

Combine terms.

Multiply $n$ by $1$ .

$\frac{- e^{n \ln (1 + \frac{1}{n})} + n e^{n \ln (1 + \frac{1}{n})} \ln (1 + \frac{1}{n}) + e^{n \ln (1 + \frac{1}{n})} \ln (1 + \frac{1}{n})}{n + \frac{1}{n} n}$

Combine $\frac{1}{n}$ and $n$ .

$\frac{- e^{n \ln (1 + \frac{1}{n})} + n e^{n \ln (1 + \frac{1}{n})} \ln (1 + \frac{1}{n}) + e^{n \ln (1 + \frac{1}{n})} \ln (1 + \frac{1}{n})}{n + \frac{n}{n}}$

Cancel the common factor of $n$ .

Cancel the common factor.

$\frac{- e^{n \ln (1 + \frac{1}{n})} + n e^{n \ln (1 + \frac{1}{n})} \ln (1 + \frac{1}{n}) + e^{n \ln (1 + \frac{1}{n})} \ln (1 + \frac{1}{n})}{n + \frac{n}{n}}$

Rewrite the expression.

$\frac{- e^{n \ln (1 + \frac{1}{n})} + n e^{n \ln (1 + \frac{1}{n})} \ln (1 + \frac{1}{n}) + e^{n \ln (1 + \frac{1}{n})} \ln (1 + \frac{1}{n})}{n + 1}$

Reorder terms.

$\frac{e^{n \ln (1 + \frac{1}{n})} n \ln (1 + \frac{1}{n}) + e^{n \ln (1 + \frac{1}{n})} \ln (1 + \frac{1}{n}) - e^{n \ln (1 + \frac{1}{n})}}{n + 1}$

Find the Derivative - d/dn (1+1/n)^n

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