Calculus Examples

$y = \arcsin (e^{x})$

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

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Step 1.1

To apply the Chain Rule, set $u$ as $e^{x}$ .

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

Step 1.2

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

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

Step 1.3

Replace all occurrences of $u$ with $e^{x}$ .

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

Step 2

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

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Step 2.1

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

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

Step 2.2

Move $2$ to the left of $x$ .

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

Step 3

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

$\frac{1}{\sqrt{1 - e^{2 x}}} e^{x}$

Step 4

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

$\frac{e^{x}}{\sqrt{1 - e^{2 x}}}$