Calculus Examples

$\arcsin (4 x) + \arccos (4 x)$

Step 1

By the Sum Rule, the derivative of $\arcsin (4 x) + \arccos (4 x)$ with respect to $x$ is $\frac{d}{d x} [\arcsin (4 x)] + \frac{d}{d x} [\arccos (4 x)]$ .

$\frac{d}{d x} [\arcsin (4 x)] + \frac{d}{d x} [\arccos (4 x)]$

Step 2

Evaluate $\frac{d}{d x} [\arcsin (4 x)]$ .

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Step 2.1

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

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Step 2.1.1

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

$\frac{d}{d u_{1}} [\arcsin (u_{1})] \frac{d}{d x} [4 x] + \frac{d}{d x} [\arccos (4 x)]$

Step 2.1.2

The derivative of $\arcsin (u_{1})$ with respect to $u_{1}$ is $\frac{1}{\sqrt{1 - {u_{1}}^{2}}}$ .

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

Step 2.1.3

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

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

Step 2.2

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

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

Step 2.3

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

$\frac{1}{\sqrt{1 - {(4 x)}^{2}}} (4 \cdot 1) + \frac{d}{d x} [\arccos (4 x)]$

Step 2.4

Factor $4$ out of $4 x$ .

$\frac{1}{\sqrt{1 - {(4 (x))}^{2}}} (4 \cdot 1) + \frac{d}{d x} [\arccos (4 x)]$

Step 2.5

Apply the product rule to $4 (x)$ .

$\frac{1}{\sqrt{1 - (4^{2} x^{2})}} (4 \cdot 1) + \frac{d}{d x} [\arccos (4 x)]$

Step 2.6

Raise $4$ to the power of $2$ .

$\frac{1}{\sqrt{1 - (16 x^{2})}} (4 \cdot 1) + \frac{d}{d x} [\arccos (4 x)]$

Step 2.7

Multiply $16$ by $- 1$ .

$\frac{1}{\sqrt{1 - 16 x^{2}}} (4 \cdot 1) + \frac{d}{d x} [\arccos (4 x)]$

Step 2.8

Multiply $4$ by $1$ .

$\frac{1}{\sqrt{1 - 16 x^{2}}} \cdot 4 + \frac{d}{d x} [\arccos (4 x)]$

Step 2.9

Combine $\frac{1}{\sqrt{1 - 16 x^{2}}}$ and $4$ .

$\frac{4}{\sqrt{1 - 16 x^{2}}} + \frac{d}{d x} [\arccos (4 x)]$

Step 3

Evaluate $\frac{d}{d x} [\arccos (4 x)]$ .

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Step 3.1

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

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Step 3.1.1

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

$\frac{4}{\sqrt{1 - 16 x^{2}}} + \frac{d}{d u_{2}} [\arccos (u_{2})] \frac{d}{d x} [4 x]$

Step 3.1.2

The derivative of $\arccos (u_{2})$ with respect to $u_{2}$ is $- \frac{1}{\sqrt{1 - {u_{2}}^{2}}}$ .

$\frac{4}{\sqrt{1 - 16 x^{2}}} - \frac{1}{\sqrt{1 - {u_{2}}^{2}}} \frac{d}{d x} [4 x]$

Step 3.1.3

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

$\frac{4}{\sqrt{1 - 16 x^{2}}} - \frac{1}{\sqrt{1 - {(4 x)}^{2}}} \frac{d}{d x} [4 x]$

Step 3.2

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$\frac{4}{\sqrt{1 - 16 x^{2}}} - \frac{1}{\sqrt{1 - {(4 x)}^{2}}} (4 \frac{d}{d x} [x])$

Step 3.3

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

$\frac{4}{\sqrt{1 - 16 x^{2}}} - \frac{1}{\sqrt{1 - {(4 x)}^{2}}} (4 \cdot 1)$

Step 3.4

Factor $4$ out of $4 x$ .

$\frac{4}{\sqrt{1 - 16 x^{2}}} - \frac{1}{\sqrt{1 - {(4 (x))}^{2}}} (4 \cdot 1)$

Step 3.5

Apply the product rule to $4 (x)$ .

$\frac{4}{\sqrt{1 - 16 x^{2}}} - \frac{1}{\sqrt{1 - (4^{2} x^{2})}} (4 \cdot 1)$

Step 3.6

Raise $4$ to the power of $2$ .

$\frac{4}{\sqrt{1 - 16 x^{2}}} - \frac{1}{\sqrt{1 - (16 x^{2})}} (4 \cdot 1)$

Step 3.7

Multiply $16$ by $- 1$ .

$\frac{4}{\sqrt{1 - 16 x^{2}}} - \frac{1}{\sqrt{1 - 16 x^{2}}} (4 \cdot 1)$

Step 3.8

Multiply $4$ by $1$ .

$\frac{4}{\sqrt{1 - 16 x^{2}}} - \frac{1}{\sqrt{1 - 16 x^{2}}} \cdot 4$

Step 3.9

Multiply $4$ by $- 1$ .

$\frac{4}{\sqrt{1 - 16 x^{2}}} - 4 \frac{1}{\sqrt{1 - 16 x^{2}}}$

Step 3.10

Combine $- 4$ and $\frac{1}{\sqrt{1 - 16 x^{2}}}$ .

$\frac{4}{\sqrt{1 - 16 x^{2}}} + \frac{- 4}{\sqrt{1 - 16 x^{2}}}$

Step 3.11

Move the negative in front of the fraction.

$\frac{4}{\sqrt{1 - 16 x^{2}}} - \frac{4}{\sqrt{1 - 16 x^{2}}}$

Step 4

Subtract $\frac{4}{\sqrt{1 - 16 x^{2}}}$ from $\frac{4}{\sqrt{1 - 16 x^{2}}}$ .

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