Calculus Examples

f (t) = 2 + e^{t}

$f (t) = 2 + e^{t}$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate.

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Step 1.1.1

By the Sum Rule, the derivative of

2 + e^{t}

$2 + e^{t}$ with respect to

t

$t$ is

\frac{d}{d t} [2] + \frac{d}{d t} [e^{t}]

$\frac{d}{d t} [2] + \frac{d}{d t} [e^{t}]$ .

\frac{d}{d t} [2] + \frac{d}{d t} [e^{t}]

$\frac{d}{d t} [2] + \frac{d}{d t} [e^{t}]$

Step 1.1.2

Since

2

$2$ is constant with respect to

t

$t$ , the derivative of

2

$2$ with respect to

t

$t$ is

0

$0$ .

0 + \frac{d}{d t} [e^{t}]

$0 + \frac{d}{d t} [e^{t}]$

0 + \frac{d}{d t} [e^{t}]

$0 + \frac{d}{d t} [e^{t}]$

Step 1.2

Differentiate using the Exponential Rule which states that

\frac{d}{d t} [a^{t}]

$\frac{d}{d t} [a^{t}]$ is

a^{t} ln (a)

$a^{t} \ln (a)$ where

a

$a$ =

e

$e$ .

0 + e^{t}

$0 + e^{t}$

Step 1.3

Add

0

$0$ and

e^{t}

$e^{t}$ .

e^{t}

$e^{t}$

e^{t}

$e^{t}$

Step 2

Differentiate using the Exponential Rule which states that

\frac{d}{d t} [a^{t}]

$\frac{d}{d t} [a^{t}]$ is

a^{t} ln (a)

$a^{t} \ln (a)$ where

a

$a$ =

e

$e$ .

$f'' (t) = e^{t}$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to

$0$ and solve.

$e^{t} = 0$

Step 4

Since there is no value of

$x$ that makes the first derivative equal to

$0$ , there are no local extrema.

No Local Extrema

Step 5

No Local Extrema