Calculus Examples

$f (y) = y \ln (1 + e^{y})$

Step 1

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

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

Step 2

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

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Step 2.1

To apply the Chain Rule, set $u$ as $1 + e^{y}$ .

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

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u$ with $1 + e^{y}$ .

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

Step 3

Differentiate.

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Step 3.1

Combine $\frac{1}{1 + e^{y}}$ and $y$ .

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

Add $0$ and $\frac{d}{d y} [e^{y}]$ .

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

Step 4

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

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

Step 5

Differentiate using the Power Rule.

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Step 5.1

Combine $\frac{y}{1 + e^{y}}$ and $e^{y}$ .

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

Step 5.2

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

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

Step 5.3

Multiply $\ln (1 + e^{y})$ by $1$ .

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

Step 6

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

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

Step 7

Combine the numerators over the common denominator.

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

Step 8

Simplify.

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Step 8.1

Simplify the numerator.

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Step 8.1.1

Simplify each term.

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Step 8.1.1.1

Apply the distributive property.

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

Step 8.1.1.2

Multiply $\ln (1 + e^{y})$ by $1$ .

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

Step 8.1.2

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

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

Step 8.2

Reorder terms.

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