Calculus Examples

$g (x) = {(x^{2} + x + 1)}^{10} \tan^{3} (x)$

Step 1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = {(x^{2} + x + 1)}^{10}$ and $g (x) = \tan^{3} (x)$ .

${(x^{2} + x + 1)}^{10} \frac{d}{d x} [\tan^{3} (x)] + \tan^{3} (x) \frac{d}{d x} [{(x^{2} + x + 1)}^{10}]$

Step 2

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

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Step 2.1

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

${(x^{2} + x + 1)}^{10} (\frac{d}{d u_{1}} [{u_{1}}^{3}] \frac{d}{d x} [\tan (x)]) + \tan^{3} (x) \frac{d}{d x} [{(x^{2} + x + 1)}^{10}]$

Step 2.2

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

${(x^{2} + x + 1)}^{10} (3 {u_{1}}^{2} \frac{d}{d x} [\tan (x)]) + \tan^{3} (x) \frac{d}{d x} [{(x^{2} + x + 1)}^{10}]$

Step 2.3

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

${(x^{2} + x + 1)}^{10} (3 \tan^{2} (x) \frac{d}{d x} [\tan (x)]) + \tan^{3} (x) \frac{d}{d x} [{(x^{2} + x + 1)}^{10}]$

Step 3

Move $3$ to the left of ${(x^{2} + x + 1)}^{10}$ .

$3 \cdot {(x^{2} + x + 1)}^{10} \tan^{2} (x) \frac{d}{d x} [\tan (x)] + \tan^{3} (x) \frac{d}{d x} [{(x^{2} + x + 1)}^{10}]$

Step 4

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

$3 {(x^{2} + x + 1)}^{10} \tan^{2} (x) \sec^{2} (x) + \tan^{3} (x) \frac{d}{d x} [{(x^{2} + x + 1)}^{10}]$

Step 5

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{10}$ and $g (x) = x^{2} + x + 1$ .

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Step 5.1

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

$3 {(x^{2} + x + 1)}^{10} \tan^{2} (x) \sec^{2} (x) + \tan^{3} (x) (\frac{d}{d u_{2}} [{u_{2}}^{10}] \frac{d}{d x} [x^{2} + x + 1])$

Step 5.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 10$ .

$3 {(x^{2} + x + 1)}^{10} \tan^{2} (x) \sec^{2} (x) + \tan^{3} (x) (10 {u_{2}}^{9} \frac{d}{d x} [x^{2} + x + 1])$

Step 5.3

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

$3 {(x^{2} + x + 1)}^{10} \tan^{2} (x) \sec^{2} (x) + \tan^{3} (x) (10 {(x^{2} + x + 1)}^{9} \frac{d}{d x} [x^{2} + x + 1])$

Step 6

Differentiate.

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Step 6.1

By the Sum Rule, the derivative of $x^{2} + x + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

$3 {(x^{2} + x + 1)}^{10} \tan^{2} (x) \sec^{2} (x) + \tan^{3} (x) (10 {(x^{2} + x + 1)}^{9} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [1]))$

Step 6.2

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

$3 {(x^{2} + x + 1)}^{10} \tan^{2} (x) \sec^{2} (x) + \tan^{3} (x) (10 {(x^{2} + x + 1)}^{9} (2 x + \frac{d}{d x} [x] + \frac{d}{d x} [1]))$

Step 6.3

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

$3 {(x^{2} + x + 1)}^{10} \tan^{2} (x) \sec^{2} (x) + \tan^{3} (x) (10 {(x^{2} + x + 1)}^{9} (2 x + 1 + \frac{d}{d x} [1]))$

Step 6.4

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

$3 {(x^{2} + x + 1)}^{10} \tan^{2} (x) \sec^{2} (x) + \tan^{3} (x) (10 {(x^{2} + x + 1)}^{9} (2 x + 1 + 0))$

Step 6.5

Simplify the expression.

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Step 6.5.1

Add $2 x + 1$ and $0$ .

$3 {(x^{2} + x + 1)}^{10} \tan^{2} (x) \sec^{2} (x) + \tan^{3} (x) (10 {(x^{2} + x + 1)}^{9} (2 x + 1))$

Step 6.5.2

Reorder terms.

$3 \sec^{2} (x) \tan^{2} (x) {(x^{2} + x + 1)}^{10} + 10 \tan^{3} (x) {(x^{2} + x + 1)}^{9} (2 x + 1)$