Calculus Examples

$y = {(2 x + 1)}^{x}$ , $x = 1$

Step 1

Use the properties of logarithms to simplify the differentiation.

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Step 1.1

Rewrite ${(2 x + 1)}^{x}$ as $e^{\ln ({(2 x + 1)}^{x})}$ .

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

Step 1.2

Expand $\ln ({(2 x + 1)}^{x})$ by moving $x$ outside the logarithm.

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

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = x \ln (2 x + 1)$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $x \ln (2 x + 1)$ .

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

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u_{1}$ with $x \ln (2 x + 1)$ .

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

Step 3

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = \ln (2 x + 1)$ .

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

Step 4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 2 x + 1$ .

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

Replace all occurrences of $u_{2}$ with $2 x + 1$ .

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

Step 5

Differentiate.

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Step 5.1

Combine $\frac{1}{2 x + 1}$ and $x$ .

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

Step 5.2

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

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

Step 5.3

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 5.4

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

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

Step 5.5

Multiply $2$ by $1$ .

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

Step 5.6

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

$e^{x \ln (2 x + 1)} (\frac{x}{2 x + 1} (2 + 0) + \ln (2 x + 1) \frac{d}{d x} [x])$

Step 5.7

Combine fractions.

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Step 5.7.1

Add $2$ and $0$ .

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

Step 5.7.2

Combine $\frac{x}{2 x + 1}$ and $2$ .

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

Step 5.7.3

Move $2$ to the left of $x$ .

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

Step 5.8

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

$e^{x \ln (2 x + 1)} (\frac{2 x}{2 x + 1} + \ln (2 x + 1) \cdot 1)$

Step 5.9

Multiply $\ln (2 x + 1)$ by $1$ .

$e^{x \ln (2 x + 1)} (\frac{2 x}{2 x + 1} + \ln (2 x + 1))$

Step 6

To write $\ln (2 x + 1)$ as a fraction with a common denominator, multiply by $\frac{2 x + 1}{2 x + 1}$ .

$e^{x \ln (2 x + 1)} (\frac{2 x}{2 x + 1} + \frac{\ln (2 x + 1) (2 x + 1)}{2 x + 1})$

Step 7

Combine the numerators over the common denominator.

$e^{x \ln (2 x + 1)} \frac{2 x + \ln (2 x + 1) (2 x + 1)}{2 x + 1}$

Step 8

Combine $e^{x \ln (2 x + 1)}$ and $\frac{2 x + \ln (2 x + 1) (2 x + 1)}{2 x + 1}$ .

$\frac{e^{x \ln (2 x + 1)} (2 x + \ln (2 x + 1) (2 x + 1))}{2 x + 1}$

Step 9

Simplify.

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Step 9.1

Simplify the numerator.

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Step 9.1.1

Simplify each term.

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Step 9.1.1.1

Apply the distributive property.

$\frac{e^{x \ln (2 x + 1)} (2 x + \ln (2 x + 1) (2 x) + \ln (2 x + 1) \cdot 1)}{2 x + 1}$

Step 9.1.1.2

Rewrite using the commutative property of multiplication.

$\frac{e^{x \ln (2 x + 1)} (2 x + 2 \ln (2 x + 1) x + \ln (2 x + 1) \cdot 1)}{2 x + 1}$

Step 9.1.1.3

Multiply $\ln (2 x + 1)$ by $1$ .

$\frac{e^{x \ln (2 x + 1)} (2 x + 2 \ln (2 x + 1) x + \ln (2 x + 1))}{2 x + 1}$

Step 9.1.1.4

Simplify $2 \ln (2 x + 1)$ by moving $2$ inside the logarithm.

$\frac{e^{x \ln (2 x + 1)} (2 x + \ln ({(2 x + 1)}^{2}) x + \ln (2 x + 1))}{2 x + 1}$

Step 9.1.2

Apply the distributive property.

$\frac{e^{x \ln (2 x + 1)} (2 x) + e^{x \ln (2 x + 1)} (\ln ({(2 x + 1)}^{2}) x) + e^{x \ln (2 x + 1)} \ln (2 x + 1)}{2 x + 1}$

Step 9.1.3

Rewrite using the commutative property of multiplication.

$\frac{2 e^{x \ln (2 x + 1)} x + e^{x \ln (2 x + 1)} (\ln ({(2 x + 1)}^{2}) x) + e^{x \ln (2 x + 1)} \ln (2 x + 1)}{2 x + 1}$

Step 9.1.4

Reorder factors in $2 e^{x \ln (2 x + 1)} x + e^{x \ln (2 x + 1)} (\ln ({(2 x + 1)}^{2}) x) + e^{x \ln (2 x + 1)} \ln (2 x + 1)$ .

$\frac{2 x e^{x \ln (2 x + 1)} + x e^{x \ln (2 x + 1)} \ln ({(2 x + 1)}^{2}) + e^{x \ln (2 x + 1)} \ln (2 x + 1)}{2 x + 1}$

Step 9.2

Reorder terms.

$\frac{2 e^{x \ln (2 x + 1)} x + e^{x \ln (2 x + 1)} x \ln ({(2 x + 1)}^{2}) + e^{x \ln (2 x + 1)} \ln (2 x + 1)}{2 x + 1}$

Step 10

Evaluate the derivative at $x = 1$ .

$\frac{2 e^{(1) \ln (2 (1) + 1)} (1) + e^{(1) \ln (2 (1) + 1)} (1) \ln ({(2 (1) + 1)}^{2}) + e^{(1) \ln (2 (1) + 1)} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11

Simplify.

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Step 11.1

Simplify the numerator.

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Step 11.1.1

Simplify $1 \ln (2 (1) + 1)$ by moving $1$ inside the logarithm.

$\frac{2 e^{\ln ({(2 (1) + 1)}^{1})} \cdot 1 + e^{1 \ln (2 (1) + 1)} \cdot 1 \ln ({(2 (1) + 1)}^{2}) + e^{1 \ln (2 (1) + 1)} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.2

Exponentiation and log are inverse functions.

$\frac{2 {(2 (1) + 1)}^{1} \cdot 1 + e^{1 \ln (2 (1) + 1)} \cdot 1 \ln ({(2 (1) + 1)}^{2}) + e^{1 \ln (2 (1) + 1)} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.3

Multiply $2$ by $1$ .

$\frac{2 {(2 + 1)}^{1} \cdot 1 + e^{1 \ln (2 (1) + 1)} \cdot 1 \ln ({(2 (1) + 1)}^{2}) + e^{1 \ln (2 (1) + 1)} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.4

Add $2$ and $1$ .

$\frac{2 \cdot 3^{1} \cdot 1 + e^{1 \ln (2 (1) + 1)} \cdot 1 \ln ({(2 (1) + 1)}^{2}) + e^{1 \ln (2 (1) + 1)} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.5

Evaluate the exponent.

$\frac{2 \cdot 3 \cdot 1 + e^{1 \ln (2 (1) + 1)} \cdot 1 \ln ({(2 (1) + 1)}^{2}) + e^{1 \ln (2 (1) + 1)} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.6

Multiply $2 \cdot 3 \cdot 1$ .

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Step 11.1.6.1

Multiply $2$ by $3$ .

$\frac{6 \cdot 1 + e^{1 \ln (2 (1) + 1)} \cdot 1 \ln ({(2 (1) + 1)}^{2}) + e^{1 \ln (2 (1) + 1)} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.6.2

Multiply $6$ by $1$ .

$\frac{6 + e^{1 \ln (2 (1) + 1)} \cdot 1 \ln ({(2 (1) + 1)}^{2}) + e^{1 \ln (2 (1) + 1)} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.7

Simplify $1 \ln (2 (1) + 1)$ by moving $1$ inside the logarithm.

$\frac{6 + e^{\ln ({(2 (1) + 1)}^{1})} \cdot 1 \ln ({(2 (1) + 1)}^{2}) + e^{1 \ln (2 (1) + 1)} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.8

Exponentiation and log are inverse functions.

$\frac{6 + {(2 (1) + 1)}^{1} \cdot 1 \ln ({(2 (1) + 1)}^{2}) + e^{1 \ln (2 (1) + 1)} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.9

Multiply $2$ by $1$ .

$\frac{6 + {(2 + 1)}^{1} \cdot 1 \ln ({(2 (1) + 1)}^{2}) + e^{1 \ln (2 (1) + 1)} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.10

Add $2$ and $1$ .

$\frac{6 + 3^{1} \cdot 1 \ln ({(2 (1) + 1)}^{2}) + e^{1 \ln (2 (1) + 1)} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.11

Evaluate the exponent.

$\frac{6 + 3 \cdot 1 \ln ({(2 (1) + 1)}^{2}) + e^{1 \ln (2 (1) + 1)} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.12

Multiply $3$ by $1$ .

$\frac{6 + 3 \ln ({(2 (1) + 1)}^{2}) + e^{1 \ln (2 (1) + 1)} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.13

Multiply $2$ by $1$ .

$\frac{6 + 3 \ln ({(2 + 1)}^{2}) + e^{1 \ln (2 (1) + 1)} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.14

Add $2$ and $1$ .

$\frac{6 + 3 \ln (3^{2}) + e^{1 \ln (2 (1) + 1)} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.15

Raise $3$ to the power of $2$ .

$\frac{6 + 3 \ln (9) + e^{1 \ln (2 (1) + 1)} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.16

Simplify $3 \ln (9)$ by moving $3$ inside the logarithm.

$\frac{6 + \ln (9^{3}) + e^{1 \ln (2 (1) + 1)} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.17

Raise $9$ to the power of $3$ .

$\frac{6 + \ln (729) + e^{1 \ln (2 (1) + 1)} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.18

Simplify $1 \ln (2 (1) + 1)$ by moving $1$ inside the logarithm.

$\frac{6 + \ln (729) + e^{\ln ({(2 (1) + 1)}^{1})} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.19

Exponentiation and log are inverse functions.

$\frac{6 + \ln (729) + {(2 (1) + 1)}^{1} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.20

Multiply $2$ by $1$ .

$\frac{6 + \ln (729) + {(2 + 1)}^{1} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.21

Add $2$ and $1$ .

$\frac{6 + \ln (729) + 3^{1} \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.22

Evaluate the exponent.

$\frac{6 + \ln (729) + 3 \ln (2 (1) + 1)}{2 (1) + 1}$

Step 11.1.23

Multiply $2$ by $1$ .

$\frac{6 + \ln (729) + 3 \ln (2 + 1)}{2 (1) + 1}$

Step 11.1.24

Add $2$ and $1$ .

$\frac{6 + \ln (729) + 3 \ln (3)}{2 (1) + 1}$

Step 11.1.25

Simplify $3 \ln (3)$ by moving $3$ inside the logarithm.

$\frac{6 + \ln (729) + \ln (3^{3})}{2 (1) + 1}$

Step 11.1.26

Raise $3$ to the power of $3$ .

$\frac{6 + \ln (729) + \ln (27)}{2 (1) + 1}$

Step 11.1.27

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\frac{6 + \ln (729 \cdot 27)}{2 (1) + 1}$

Step 11.1.28

Multiply $729$ by $27$ .

$\frac{6 + \ln (19683)}{2 (1) + 1}$

Step 11.2

Simplify the denominator.

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Step 11.2.1

Multiply $2$ by $1$ .

$\frac{6 + \ln (19683)}{2 + 1}$

Step 11.2.2

Add $2$ and $1$ .

$\frac{6 + \ln (19683)}{3}$