Calculus Examples

$\tan (\ln (a x + b))$

Step 1

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

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Step 1.1

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

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

Step 1.2

The derivative of $\tan (u_{1})$ with respect to $u_{1}$ is $\sec^{2} (u_{1})$ .

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

Step 1.3

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

$\sec^{2} (\ln (a x + b)) \frac{d}{d a} [\ln (a x + b)]$

Step 2

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 $g (a) = a x + b$ .

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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

Step 3

Differentiate.

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Step 3.1

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

$\frac{\sec^{2} (\ln (a x + b))}{a x + b} \frac{d}{d a} [a x + b]$

Step 3.2

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

$\frac{\sec^{2} (\ln (a x + b))}{a x + b} (\frac{d}{d a} [a x] + \frac{d}{d a} [b])$

Step 3.3

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

$\frac{\sec^{2} (\ln (a x + b))}{a x + b} (x \frac{d}{d a} [a] + \frac{d}{d a} [b])$

Step 3.4

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

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

Step 3.5

Multiply $x$ by $1$ .

$\frac{\sec^{2} (\ln (a x + b))}{a x + b} (x + \frac{d}{d a} [b])$

Step 3.6

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

$\frac{\sec^{2} (\ln (a x + b))}{a x + b} (x + 0)$

Step 3.7

Combine fractions.

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Step 3.7.1

Add $x$ and $0$ .

$\frac{\sec^{2} (\ln (a x + b))}{a x + b} x$

Step 3.7.2

Combine $\frac{\sec^{2} (\ln (a x + b))}{a x + b}$ and $x$ .

$\frac{\sec^{2} (\ln (a x + b)) x}{a x + b}$

Step 4

Simplify.

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Step 4.1

Reorder factors in $\sec^{2} (\ln (a x + b)) x$ .

$\frac{x \sec^{2} (\ln (a x + b))}{a x + b}$

Step 4.2

Reorder terms.

$\frac{x \sec^{2} (\ln (a x + b))}{b + a x}$