Calculus Examples

$1 - \arctan (e^{t})$

Step 1

Differentiate.

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Step 1.1

By the Sum Rule, the derivative of $1 - \arctan (e^{t})$ with respect to $t$ is $\frac{d}{d t} [1] + \frac{d}{d t} [- \arctan (e^{t})]$ .

$\frac{d}{d t} [1] + \frac{d}{d t} [- \arctan (e^{t})]$

Step 1.2

Since $1$ is constant with respect to $t$ , the derivative of $1$ with respect to $t$ is $0$ .

$0 + \frac{d}{d t} [- \arctan (e^{t})]$

Step 2

Evaluate $\frac{d}{d t} [- \arctan (e^{t})]$ .

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Step 2.1

Since $- 1$ is constant with respect to $t$ , the derivative of $- \arctan (e^{t})$ with respect to $t$ is $- \frac{d}{d t} [\arctan (e^{t})]$ .

$0 - \frac{d}{d t} [\arctan (e^{t})]$

Step 2.2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \arctan (t)$ and $g (t) = e^{t}$ .

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Step 2.2.1

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

$0 - (\frac{d}{d u} [\arctan (u)] \frac{d}{d t} [e^{t}])$

Step 2.2.2

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

$0 - (\frac{1}{1 + u^{2}} \frac{d}{d t} [e^{t}])$

Step 2.2.3

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

$0 - (\frac{1}{1 + {(e^{t})}^{2}} \frac{d}{d t} [e^{t}])$

Step 2.3

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

$0 - (\frac{1}{1 + {(e^{t})}^{2}} e^{t})$

Step 2.4

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

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Step 2.4.1

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

$0 - (\frac{1}{1 + e^{t \cdot 2}} e^{t})$

Step 2.4.2

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

$0 - (\frac{1}{1 + e^{2 t}} e^{t})$

Step 2.5

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

$0 - \frac{e^{t}}{1 + e^{2 t}}$

Step 3

Subtract $\frac{e^{t}}{1 + e^{2 t}}$ from $0$ .

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