Calculus Examples

$\arccos (e^{- t})$

Step 1

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

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Step 1.1

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

$\frac{d}{d u_{1}} [\arccos (u_{1})] \frac{d}{d t} [e^{- t}]$

Step 1.2

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

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

Step 1.3

Replace all occurrences of $u_{1}$ with $e^{- t}$ .

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

Step 2

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

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Step 2.1

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

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

Step 2.2

Multiply $2$ by $- 1$ .

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

Step 3

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

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Step 3.1

To apply the Chain Rule, set $u_{2}$ as $- t$ .

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

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 $a$ = $e$ .

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

Step 3.3

Replace all occurrences of $u_{2}$ with $- t$ .

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

Step 4

Differentiate.

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

Multiply.

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Step 4.3.1

Multiply $- 1$ by $- 1$ .

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

Step 4.3.2

Multiply $\frac{e^{- t}}{\sqrt{1 - e^{- 2 t}}}$ by $1$ .

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

Step 4.4

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

$\frac{e^{- t}}{\sqrt{1 - e^{- 2 t}}} \cdot 1$

Step 4.5

Multiply $\frac{e^{- t}}{\sqrt{1 - e^{- 2 t}}}$ by $1$ .

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