Calculus Examples

$P (t) = 2 {(8)}^{0.019 t}$

Step 1

Rewrite $8$ as $2^{3}$ .

$\frac{d}{d t} [2 {(2^{3})}^{0.019 t}]$

Step 2

Multiply the exponents in ${(2^{3})}^{0.019 t}$ .

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Step 2.1

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

$\frac{d}{d t} [2 \cdot 2^{3 (0.019 t)}]$

Step 2.2

Multiply $0.019$ by $3$ .

$\frac{d}{d t} [2 \cdot 2^{0.057 t}]$

Step 3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{d}{d t} [2^{1 + 0.057 t}]$

Step 4

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = 2^{t}$ and $g (t) = 1 + 0.057 t$ .

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Step 4.1

To apply the Chain Rule, set $u$ as $1 + 0.057 t$ .

$\frac{d}{d u} [2^{u}] \frac{d}{d t} [1 + 0.057 t]$

Step 4.2

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

$2^{u} \ln (2) \frac{d}{d t} [1 + 0.057 t]$

Step 4.3

Replace all occurrences of $u$ with $1 + 0.057 t$ .

$2^{1 + 0.057 t} \ln (2) \frac{d}{d t} [1 + 0.057 t]$

Step 5

Differentiate.

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Step 5.1

By the Sum Rule, the derivative of $1 + 0.057 t$ with respect to $t$ is $\frac{d}{d t} [1] + \frac{d}{d t} [0.057 t]$ .

$2^{1 + 0.057 t} \ln (2) (\frac{d}{d t} [1] + \frac{d}{d t} [0.057 t])$

Step 5.2

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

$2^{1 + 0.057 t} \ln (2) (0 + \frac{d}{d t} [0.057 t])$

Step 5.3

Add $0$ and $\frac{d}{d t} [0.057 t]$ .

$2^{1 + 0.057 t} \ln (2) \frac{d}{d t} [0.057 t]$

Step 5.4

Since $0.057$ is constant with respect to $t$ , the derivative of $0.057 t$ with respect to $t$ is $0.057 \frac{d}{d t} [t]$ .

$2^{1 + 0.057 t} \ln (2) (0.057 \frac{d}{d t} [t])$

Step 5.5

Simplify by multiplying terms.

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Step 5.5.1

Multiply $0.057$ by $\ln (2)$ .

$2^{1 + 0.057 t} \cdot 0.03950938 \frac{d}{d t} [t]$

Step 5.5.2

Move $0.03950938$ to the left of $2^{1 + 0.057 t}$ .

$0.03950938 \cdot 2^{1 + 0.057 t} \frac{d}{d t} [t]$

Step 5.6

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

$0.03950938 \cdot 2^{1 + 0.057 t} \cdot 1$

Step 5.7

Multiply $0.03950938$ by $1$ .

$0.03950938 \cdot 2^{1 + 0.057 t}$