Calculus Examples

$s = 2 \cot (t + e^{- t})$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d t} (s) = \frac{d}{d t} (2 \cot (t + e^{- t}))$

Step 2

The derivative of $s$ with respect to $t$ is $s'$ .

$s'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Since $2$ is constant with respect to $t$ , the derivative of $2 \cot (t + e^{- t})$ with respect to $t$ is $2 \frac{d}{d t} [\cot (t + e^{- t})]$ .

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

The derivative of $\cot (u_{1})$ with respect to $u_{1}$ is $- \csc^{2} (u_{1})$ .

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

Step 3.2.3

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

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

Step 3.3

Differentiate.

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Step 3.3.1

Multiply $- 1$ by $2$ .

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.4

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 3.4.1

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

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

Step 3.4.2

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

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

Step 3.4.3

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

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

Step 3.5

Differentiate.

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Step 3.5.1

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

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

Step 3.5.2

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

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

Step 3.5.3

Simplify the expression.

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Step 3.5.3.1

Multiply $- 1$ by $1$ .

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

Step 3.5.3.2

Move $- 1$ to the left of $e^{- t}$ .

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

Step 3.5.3.3

Rewrite $- 1 e^{- t}$ as $- e^{- t}$ .

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

Step 4

Reform the equation by setting the left side equal to the right side.

$s' = - 2 \csc^{2} (t + e^{- t}) (1 - e^{- t})$

Step 5

Replace $s'$ with $\frac{d s}{d t}$ .

$\frac{d s}{d t} = - 2 \csc^{2} (t + e^{- t}) (1 - e^{- t})$