Calculus Examples

$arccot (e^{s})$

Step 1

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

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

$\frac{d}{d u} [arccot (u)] \frac{d}{d s} [e^{s}]$

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

$- \frac{1}{1 + u^{2}} \frac{d}{d s} [e^{s}]$

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

$- \frac{1}{1 + {(e^{s})}^{2}} \frac{d}{d s} [e^{s}]$

Step 2

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

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

$- \frac{1}{1 + e^{s \cdot 2}} \frac{d}{d s} [e^{s}]$

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

$- \frac{1}{1 + e^{2 s}} \frac{d}{d s} [e^{s}]$

Step 3

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

$- \frac{1}{1 + e^{2 s}} e^{s}$

Step 4

Combine $e^{s}$ and $\frac{1}{1 + e^{2 s}}$ .

$- \frac{e^{s}}{1 + e^{2 s}}$