Calculus Examples

$\arcsin (e^{3 x})$

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 $g (x) = e^{3 x}$ .

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $e^{3 x}$ .

$\frac{d}{d u_{1}} [\arcsin (u_{1})] \frac{d}{d x} [e^{3 x}]$

Step 1.2

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

$\frac{1}{\sqrt{1 - {u_{1}}^{2}}} \frac{d}{d x} [e^{3 x}]$

Step 1.3

Replace all occurrences of $u_{1}$ with $e^{3 x}$ .

$\frac{1}{\sqrt{1 - {(e^{3 x})}^{2}}} \frac{d}{d x} [e^{3 x}]$

Step 2

Multiply the exponents in ${(e^{3 x})}^{2}$ .

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Step 2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{1}{\sqrt{1 - e^{3 x \cdot 2}}} \frac{d}{d x} [e^{3 x}]$

Step 2.2

Multiply $2$ by $3$ .

$\frac{1}{\sqrt{1 - e^{6 x}}} \frac{d}{d x} [e^{3 x}]$

Step 3

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

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Step 3.1

To apply the Chain Rule, set $u_{2}$ as $3 x$ .

$\frac{1}{\sqrt{1 - e^{6 x}}} (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [3 x])$

Step 3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$\frac{1}{\sqrt{1 - e^{6 x}}} (e^{u_{2}} \frac{d}{d x} [3 x])$

Step 3.3

Replace all occurrences of $u_{2}$ with $3 x$ .

$\frac{1}{\sqrt{1 - e^{6 x}}} (e^{3 x} \frac{d}{d x} [3 x])$

Step 4

Differentiate.

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Step 4.1

Combine $e^{3 x}$ and $\frac{1}{\sqrt{1 - e^{6 x}}}$ .

$\frac{e^{3 x}}{\sqrt{1 - e^{6 x}}} \frac{d}{d x} [3 x]$

Step 4.2

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

$\frac{e^{3 x}}{\sqrt{1 - e^{6 x}}} (3 \frac{d}{d x} [x])$

Step 4.3

Combine $3$ and $\frac{e^{3 x}}{\sqrt{1 - e^{6 x}}}$ .

$\frac{3 e^{3 x}}{\sqrt{1 - e^{6 x}}} \frac{d}{d x} [x]$

Step 4.4

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

$\frac{3 e^{3 x}}{\sqrt{1 - e^{6 x}}} \cdot 1$

Step 4.5

Multiply $\frac{3 e^{3 x}}{\sqrt{1 - e^{6 x}}}$ by $1$ .

$\frac{3 e^{3 x}}{\sqrt{1 - e^{6 x}}}$