Calculus Examples

y = ln (e^{x} + x e^{x})

$y = \ln (e^{x} + x e^{x})$

Step 1

Let

y = f (x)

$y = f (x)$ , take the natural logarithm of both sides

ln (y) = ln (f (x))

$\ln (y) = \ln (f (x))$ .

ln (y) = ln (ln (e^{x} + x e^{x}))

$\ln (y) = \ln (\ln (e^{x} + x e^{x}))$

Step 2

Differentiate the expression using the chain rule, keeping in mind that

y

$y$ is a function of

x

$x$ .

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Step 2.1

Differentiate the left hand side

ln (y)

$\ln (y)$ using the chain rule.

$\frac{y'}{y} = \ln (\ln (e^{x} + x e^{x}))$

Step 2.2

Differentiate the right hand side.

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Step 2.2.1

Differentiate

$\ln (\ln (e^{x} + x e^{x}))$ .

$\frac{y'}{y} = \frac{d}{d x} [\ln (\ln (e^{x} + x e^{x}))]$

Step 2.2.2

Differentiate using the chain rule, which states that

$\frac{d}{d x} [f (g (x))]$ is

$f' (g (x)) g' (x)$ where

$f (x) = \ln (x)$ and

$g (x) = \ln (e^{x} + x e^{x})$ .

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Step 2.2.2.1

To apply the Chain Rule, set

$u_{1}$ as

$\ln (e^{x} + x e^{x})$ .

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

Step 2.2.2.2

The derivative of

$\ln (u_{1})$ with respect to

$u_{1}$ is

$\frac{1}{u_{1}}$ .

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

Step 2.2.2.3

Replace all occurrences of

$u_{1}$ with

$\ln (e^{x} + x e^{x})$ .

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

Step 2.2.3

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

$g (x) = e^{x} + x e^{x}$ .

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Step 2.2.3.1

To apply the Chain Rule, set

$u_{2}$ as

$e^{x} + x e^{x}$ .

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

Step 2.2.3.2

The derivative of

$\ln (u_{2})$ with respect to

$u_{2}$ is

$\frac{1}{u_{2}}$ .

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

Step 2.2.3.3

Replace all occurrences of

$u_{2}$ with

$e^{x} + x e^{x}$ .

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

Step 2.2.4

Differentiate using the Sum Rule.

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Step 2.2.4.1

Multiply

$\frac{1}{e^{x} + x e^{x}}$ by

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

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

Step 2.2.4.2

By the Sum Rule, the derivative of

$e^{x} + x e^{x}$ with respect to

$x$ is

$\frac{d}{d x} [e^{x}] + \frac{d}{d x} [x e^{x}]$ .

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

Step 2.2.5

Differentiate using the Exponential Rule which states that

$\frac{d}{d x} [a^{x}]$ is

$a^{x} \ln (a)$ where

$a$ =

$e$ .

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

Step 2.2.6

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

$g (x) = e^{x}$ .

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

Step 2.2.7

Differentiate using the Exponential Rule which states that

$\frac{d}{d x} [a^{x}]$ is

$a^{x} \ln (a)$ where

$a$ =

$e$ .

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

Step 2.2.8

Differentiate using the Power Rule.

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Step 2.2.8.1

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

$\frac{y'}{y} = \frac{1}{(e^{x} + x e^{x}) \ln (e^{x} + x e^{x})} (e^{x} + x e^{x} + e^{x} \cdot 1)$

Step 2.2.8.2

Simplify by adding terms.

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Step 2.2.8.2.1

Multiply

$e^{x}$ by

$1$ .

$\frac{y'}{y} = \frac{1}{(e^{x} + x e^{x}) \ln (e^{x} + x e^{x})} (e^{x} + x e^{x} + e^{x})$

Step 2.2.8.2.2

Add

$e^{x}$ and

$e^{x}$ .

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

Step 2.2.9

Simplify.

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Step 2.2.9.1

Reorder the factors of

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

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

Step 2.2.9.2

Factor

$e^{x}$ out of

$e^{x} + x e^{x}$ .

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Step 2.2.9.2.1

Multiply by

$1$ .

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

Step 2.2.9.2.2

Factor

$e^{x}$ out of

$x e^{x}$ .

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

Step 2.2.9.2.3

Factor

$e^{x}$ out of

$e^{x} \cdot 1 + e^{x} x$ .

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

Step 2.2.9.3

Multiply

$x e^{x} + 2 e^{x}$ by

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

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

Step 2.2.9.4

Factor

$e^{x}$ out of

$x e^{x} + 2 e^{x}$ .

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Step 2.2.9.4.1

Factor

$e^{x}$ out of

$x e^{x}$ .

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

Step 2.2.9.4.2

Factor

$e^{x}$ out of

$2 e^{x}$ .

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

Step 2.2.9.4.3

Factor

$e^{x}$ out of

$e^{x} x + e^{x} \cdot 2$ .

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

Step 2.2.9.5

Cancel the common factor of

$e^{x}$ .

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Step 2.2.9.5.1

Cancel the common factor.

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

Step 2.2.9.5.2

Rewrite the expression.

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

Step 2.2.9.6

Reorder factors in

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

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

Step 3

Isolate

$y'$ and substitute the original function for

$y$ in the right hand side.

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

Step 4

Cancel the common factor of

$\ln (e^{x} + x e^{x})$ .

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Step 4.1

Cancel the common factor.

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

Step 4.2

Rewrite the expression.

$y' = \frac{x + 2}{1 + x}$