Calculus Examples

$\tan^{2} (x^{4} + 1)$

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 $g (x) = \tan (x^{4} + 1)$ .

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $\tan (x^{4} + 1)$ .

$\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d x} [\tan (x^{4} + 1)]$

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 $n = 2$ .

$2 u_{1} \frac{d}{d x} [\tan (x^{4} + 1)]$

Step 1.3

Replace all occurrences of $u_{1}$ with $\tan (x^{4} + 1)$ .

$2 \tan (x^{4} + 1) \frac{d}{d x} [\tan (x^{4} + 1)]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \tan (x)$ and $g (x) = x^{4} + 1$ .

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Step 2.1

To apply the Chain Rule, set $u_{2}$ as $x^{4} + 1$ .

$2 \tan (x^{4} + 1) (\frac{d}{d u_{2}} [\tan (u_{2})] \frac{d}{d x} [x^{4} + 1])$

Step 2.2

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

$2 \tan (x^{4} + 1) (\sec^{2} (u_{2}) \frac{d}{d x} [x^{4} + 1])$

Step 2.3

Replace all occurrences of $u_{2}$ with $x^{4} + 1$ .

$2 \tan (x^{4} + 1) (\sec^{2} (x^{4} + 1) \frac{d}{d x} [x^{4} + 1])$

Step 3

Differentiate.

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Step 3.1

By the Sum Rule, the derivative of $x^{4} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [1]$ .

$2 \tan (x^{4} + 1) \sec^{2} (x^{4} + 1) (\frac{d}{d x} [x^{4}] + \frac{d}{d x} [1])$

Step 3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$2 \tan (x^{4} + 1) \sec^{2} (x^{4} + 1) (4 x^{3} + \frac{d}{d x} [1])$

Step 3.3

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

$2 \tan (x^{4} + 1) \sec^{2} (x^{4} + 1) (4 x^{3} + 0)$

Step 3.4

Simplify the expression.

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Step 3.4.1

Add $4 x^{3}$ and $0$ .

$2 \tan (x^{4} + 1) \sec^{2} (x^{4} + 1) (4 x^{3})$

Step 3.4.2

Multiply $4$ by $2$ .

$8 \tan (x^{4} + 1) \sec^{2} (x^{4} + 1) x^{3}$

Step 3.4.3

Reorder the factors of $8 \tan (x^{4} + 1) \sec^{2} (x^{4} + 1) x^{3}$ .

$8 x^{3} \sec^{2} (x^{4} + 1) \tan (x^{4} + 1)$