Calculus Examples

$\arcsin (4 x)$

Step 1

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

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Step 1.1

To apply the Chain Rule, set $u$ as $4 x$ .

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

Step 1.2

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

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

Step 1.3

Replace all occurrences of $u$ with $4 x$ .

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

Step 2

Differentiate.

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Step 2.1

Factor $4$ out of $4 x$ .

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

Step 2.2

Simplify the expression.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Multiply $16$ by $- 1$ .

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

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

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

Step 2.6

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

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