Find the Derivative - d/d@VAR f(t)=1/2e^(3t-1)

$f (t) = \frac{1}{2} e^{3 t - 1}$

Step 1

Since $\frac{1}{2}$ is constant with respect to $t$ , the derivative of $\frac{1}{2} \cdot e^{3 t - 1}$ with respect to $t$ is $\frac{1}{2} \frac{d}{d t} [e^{3 t - 1}]$ .

$\frac{1}{2} \frac{d}{d t} [e^{3 t - 1}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = 3 t - 1$ .

To apply the Chain Rule, set $u$ as $3 t - 1$ .

$\frac{1}{2} (\frac{d}{d u} [e^{u}] \frac{d}{d t} [3 t - 1])$

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

$\frac{1}{2} (e^{u} \frac{d}{d t} [3 t - 1])$

Replace all occurrences of $u$ with $3 t - 1$ .

$\frac{1}{2} (e^{3 t - 1} \frac{d}{d t} [3 t - 1])$

Step 3

Differentiate.

Combine $e^{3 t - 1}$ and $\frac{1}{2}$ .

$\frac{e^{3 t - 1}}{2} \frac{d}{d t} [3 t - 1]$

By the Sum Rule, the derivative of $3 t - 1$ with respect to $t$ is $\frac{d}{d t} [3 t] + \frac{d}{d t} [- 1]$ .

$\frac{e^{3 t - 1}}{2} (\frac{d}{d t} [3 t] + \frac{d}{d t} [- 1])$

Since $3$ is constant with respect to $t$ , the derivative of $3 t$ with respect to $t$ is $3 \frac{d}{d t} [t]$ .

$\frac{e^{3 t - 1}}{2} (3 \frac{d}{d t} [t] + \frac{d}{d t} [- 1])$

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

$\frac{e^{3 t - 1}}{2} (3 \cdot 1 + \frac{d}{d t} [- 1])$

Multiply $3$ by $1$ .

$\frac{e^{3 t - 1}}{2} (3 + \frac{d}{d t} [- 1])$

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

$\frac{e^{3 t - 1}}{2} (3 + 0)$

Combine fractions.

Add $3$ and $0$ .

$\frac{e^{3 t - 1}}{2} \cdot 3$

Combine $\frac{e^{3 t - 1}}{2}$ and $3$ .

$\frac{e^{3 t - 1} \cdot 3}{2}$

Move $3$ to the left of $e^{3 t - 1}$ .

$\frac{3 e^{3 t - 1}}{2}$

Find the Derivative - d/d@VAR f(t)=1/2e^(3t-1)

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