Calculus Examples

$j (x) = \ln (e^{a x} + b)$

Step 1

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

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $e^{a x} + b$ .

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

Step 1.2

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

$\frac{1}{u_{1}} \frac{d}{d a} [e^{a x} + b]$

Step 1.3

Replace all occurrences of $u_{1}$ with $e^{a x} + b$ .

$\frac{1}{e^{a x} + b} \frac{d}{d a} [e^{a x} + b]$

Step 2

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

$\frac{1}{e^{a x} + b} (\frac{d}{d a} [e^{a x}] + \frac{d}{d a} [b])$

Step 3

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

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Step 3.1

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

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

Step 3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

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

Step 3.3

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

$\frac{1}{e^{a x} + b} (e^{a x} \frac{d}{d a} [a x] + \frac{d}{d a} [b])$

Step 4

Differentiate.

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Step 4.1

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

$\frac{1}{e^{a x} + b} (e^{a x} (x \frac{d}{d a} [a]) + \frac{d}{d a} [b])$

Step 4.2

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

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

Step 4.3

Multiply $x$ by $1$ .

$\frac{1}{e^{a x} + b} (e^{a x} x + \frac{d}{d a} [b])$

Step 4.4

Since $b$ is constant with respect to $a$ , the derivative of $b$ with respect to $a$ is $0$ .

$\frac{1}{e^{a x} + b} (e^{a x} x + 0)$

Step 4.5

Combine fractions.

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Step 4.5.1

Add $e^{a x} x$ and $0$ .

$\frac{1}{e^{a x} + b} (e^{a x} x)$

Step 4.5.2

Combine $e^{a x}$ and $\frac{1}{e^{a x} + b}$ .

$\frac{e^{a x}}{e^{a x} + b} x$

Step 4.5.3

Combine $\frac{e^{a x}}{e^{a x} + b}$ and $x$ .

$\frac{e^{a x} x}{e^{a x} + b}$

Step 5

Simplify.

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Step 5.1

Reorder factors in $e^{a x} x$ .

$\frac{x e^{a x}}{e^{a x} + b}$

Step 5.2

Reorder terms.

$\frac{e^{a x} x}{b + e^{a x}}$

Step 5.3

Reorder factors in $\frac{e^{a x} x}{b + e^{a x}}$ .

$\frac{x e^{a x}}{b + e^{a x}}$