Calculus Examples

$y = \tanh^{2} ({(3 x + 2)}^{2})$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \tanh ({(3 x + 2)}^{2})$ .

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $\tanh ({(3 x + 2)}^{2})$ .

$\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d x} [\tanh ({(3 x + 2)}^{2})]$

Step 1.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 2$ .

$2 u_{1} \frac{d}{d x} [\tanh ({(3 x + 2)}^{2})]$

Step 1.3

Replace all occurrences of $u_{1}$ with $\tanh ({(3 x + 2)}^{2})$ .

$2 \tanh ({(3 x + 2)}^{2}) \frac{d}{d x} [\tanh ({(3 x + 2)}^{2})]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \tanh (x)$ and $g (x) = {(3 x + 2)}^{2}$ .

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Step 2.1

To apply the Chain Rule, set $u_{2}$ as ${(3 x + 2)}^{2}$ .

$2 \tanh ({(3 x + 2)}^{2}) (\frac{d}{d u_{2}} [\tanh (u_{2})] \frac{d}{d x} [{(3 x + 2)}^{2}])$

Step 2.2

The derivative of $\tanh (u_{2})$ with respect to $u_{2}$ is ${sech}^{2} (u_{2})$ .

$2 \tanh ({(3 x + 2)}^{2}) ({sech}^{2} (u_{2}) \frac{d}{d x} [{(3 x + 2)}^{2}])$

Step 2.3

Replace all occurrences of $u_{2}$ with ${(3 x + 2)}^{2}$ .

$2 \tanh ({(3 x + 2)}^{2}) ({sech}^{2} ({(3 x + 2)}^{2}) \frac{d}{d x} [{(3 x + 2)}^{2}])$

Step 3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = 3 x + 2$ .

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Step 3.1

To apply the Chain Rule, set $u_{3}$ as $3 x + 2$ .

$2 \tanh ({(3 x + 2)}^{2}) ({sech}^{2} ({(3 x + 2)}^{2}) (\frac{d}{d u_{3}} [{u_{3}}^{2}] \frac{d}{d x} [3 x + 2]))$

Step 3.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{3}} [{u_{3}}^{n}]$ is $n {u_{3}}^{n - 1}$ where $n = 2$ .

$2 \tanh ({(3 x + 2)}^{2}) ({sech}^{2} ({(3 x + 2)}^{2}) (2 u_{3} \frac{d}{d x} [3 x + 2]))$

Step 3.3

Replace all occurrences of $u_{3}$ with $3 x + 2$ .

$2 \tanh ({(3 x + 2)}^{2}) ({sech}^{2} ({(3 x + 2)}^{2}) (2 (3 x + 2) \frac{d}{d x} [3 x + 2]))$

Step 4

Differentiate.

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Step 4.1

Multiply $2$ by $2$ .

$4 \tanh ({(3 x + 2)}^{2}) ({sech}^{2} ({(3 x + 2)}^{2}) ((3 x + 2) \frac{d}{d x} [3 x + 2]))$

Step 4.2

By the Sum Rule, the derivative of $3 x + 2$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [2]$ .

$4 \tanh ({(3 x + 2)}^{2}) ({sech}^{2} ({(3 x + 2)}^{2}) ((3 x + 2) (\frac{d}{d x} [3 x] + \frac{d}{d x} [2])))$

Step 4.3

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$4 \tanh ({(3 x + 2)}^{2}) ({sech}^{2} ({(3 x + 2)}^{2}) ((3 x + 2) (3 \frac{d}{d x} [x] + \frac{d}{d x} [2])))$

Step 4.4

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

$4 \tanh ({(3 x + 2)}^{2}) ({sech}^{2} ({(3 x + 2)}^{2}) ((3 x + 2) (3 \cdot 1 + \frac{d}{d x} [2])))$

Step 4.5

Multiply $3$ by $1$ .

$4 \tanh ({(3 x + 2)}^{2}) ({sech}^{2} ({(3 x + 2)}^{2}) ((3 x + 2) (3 + \frac{d}{d x} [2])))$

Step 4.6

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$4 \tanh ({(3 x + 2)}^{2}) ({sech}^{2} ({(3 x + 2)}^{2}) ((3 x + 2) (3 + 0)))$

Step 4.7

Simplify the expression.

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Step 4.7.1

Add $3$ and $0$ .

$4 \tanh ({(3 x + 2)}^{2}) ({sech}^{2} ({(3 x + 2)}^{2}) ((3 x + 2) \cdot 3))$

Step 4.7.2

Move $3$ to the left of $3 x + 2$ .

$4 \tanh ({(3 x + 2)}^{2}) ({sech}^{2} ({(3 x + 2)}^{2}) (3 \cdot (3 x + 2)))$

Step 4.7.3

Multiply $3$ by $4$ .

$12 \tanh ({(3 x + 2)}^{2}) ({sech}^{2} ({(3 x + 2)}^{2}) (3 x + 2))$

Step 4.7.4

Reorder the factors of $12 \tanh ({(3 x + 2)}^{2}) ({sech}^{2} ({(3 x + 2)}^{2}) (3 x + 2))$ .

$12 {sech}^{2} ({(3 x + 2)}^{2}) \tanh ({(3 x + 2)}^{2}) (3 x + 2)$