Calculus Examples

$y = 4^{\sqrt{t}}$

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{t}$ as $t^{\frac{1}{2}}$ .

$\frac{d}{d t} [4^{t^{\frac{1}{2}}}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = 4^{t}$ and $g (t) = t^{\frac{1}{2}}$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $t^{\frac{1}{2}}$ .

$\frac{d}{d u} [4^{u}] \frac{d}{d t} [t^{\frac{1}{2}}]$

Step 2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $4$ .

$4^{u} \ln (4) \frac{d}{d t} [t^{\frac{1}{2}}]$

Step 2.3

Replace all occurrences of $u$ with $t^{\frac{1}{2}}$ .

$4^{t^{\frac{1}{2}}} \ln (4) \frac{d}{d t} [t^{\frac{1}{2}}]$

Step 3

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = \frac{1}{2}$ .

$4^{t^{\frac{1}{2}}} \ln (4) (\frac{1}{2} t^{\frac{1}{2} - 1})$

Step 4

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$4^{t^{\frac{1}{2}}} \ln (4) (\frac{1}{2} t^{\frac{1}{2} - 1 \cdot \frac{2}{2}})$

Step 5

Combine $- 1$ and $\frac{2}{2}$ .

$4^{t^{\frac{1}{2}}} \ln (4) (\frac{1}{2} t^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}})$

Step 6

Combine the numerators over the common denominator.

$4^{t^{\frac{1}{2}}} \ln (4) (\frac{1}{2} t^{\frac{1 - 1 \cdot 2}{2}})$

Step 7

Simplify the numerator.

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Step 7.1

Multiply $- 1$ by $2$ .

$4^{t^{\frac{1}{2}}} \ln (4) (\frac{1}{2} t^{\frac{1 - 2}{2}})$

Step 7.2

Subtract $2$ from $1$ .

$4^{t^{\frac{1}{2}}} \ln (4) (\frac{1}{2} t^{\frac{- 1}{2}})$

Step 8

Move the negative in front of the fraction.

$4^{t^{\frac{1}{2}}} \ln (4) (\frac{1}{2} t^{- \frac{1}{2}})$

Step 9

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

$4^{t^{\frac{1}{2}}} \ln (4) \frac{t^{- \frac{1}{2}}}{2}$

Step 10

Combine $\frac{t^{- \frac{1}{2}}}{2}$ and $4^{t^{\frac{1}{2}}}$ .

$\frac{t^{- \frac{1}{2}} \cdot 4^{t^{\frac{1}{2}}}}{2} \ln (4)$

Step 11

Combine $\frac{t^{- \frac{1}{2}} \cdot 4^{t^{\frac{1}{2}}}}{2}$ and $\ln (4)$ .

$\frac{t^{- \frac{1}{2}} \cdot 4^{t^{\frac{1}{2}}} \ln (4)}{2}$

Step 12

Move $t^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{4^{t^{\frac{1}{2}}} \ln (4)}{2 t^{\frac{1}{2}}}$