Calculus Examples

$6 \tan (3 y + 1)$

Step 1

Since $6$ is constant with respect to $y$ , the derivative of $6 \tan (3 y + 1)$ with respect to $y$ is $6 \frac{d}{d y} [\tan (3 y + 1)]$ .

$6 \frac{d}{d y} [\tan (3 y + 1)]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = \tan (y)$ and $g (y) = 3 y + 1$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $3 y + 1$ .

$6 (\frac{d}{d u} [\tan (u)] \frac{d}{d y} [3 y + 1])$

Step 2.2

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

$6 (\sec^{2} (u) \frac{d}{d y} [3 y + 1])$

Step 2.3

Replace all occurrences of $u$ with $3 y + 1$ .

$6 (\sec^{2} (3 y + 1) \frac{d}{d y} [3 y + 1])$

Step 3

Differentiate.

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Step 3.1

By the Sum Rule, the derivative of $3 y + 1$ with respect to $y$ is $\frac{d}{d y} [3 y] + \frac{d}{d y} [1]$ .

$6 \sec^{2} (3 y + 1) (\frac{d}{d y} [3 y] + \frac{d}{d y} [1])$

Step 3.2

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

$6 \sec^{2} (3 y + 1) (3 \frac{d}{d y} [y] + \frac{d}{d y} [1])$

Step 3.3

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

$6 \sec^{2} (3 y + 1) (3 \cdot 1 + \frac{d}{d y} [1])$

Step 3.4

Multiply $3$ by $1$ .

$6 \sec^{2} (3 y + 1) (3 + \frac{d}{d y} [1])$

Step 3.5

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

$6 \sec^{2} (3 y + 1) (3 + 0)$

Step 3.6

Simplify the expression.

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Step 3.6.1

Add $3$ and $0$ .

$6 \sec^{2} (3 y + 1) \cdot 3$

Step 3.6.2

Multiply $3$ by $6$ .

$18 \sec^{2} (3 y + 1)$