Calculus Examples

$f (x) = \frac{t + 1}{2^{t}}$

Step 1

Differentiate using the Quotient Rule which states that $\frac{d}{d t} [\frac{f (t)}{g (t)}]$ is $\frac{g (t) \frac{d}{d t} [f (t)] - f (t) \frac{d}{d t} [g (t)]}{{g (t)}^{2}}$ where $f (t) = t + 1$ and $g (t) = 2^{t}$ .

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

Step 2

Differentiate.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

Move $2$ to the left of $t$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

Simplify the expression.

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Step 2.5.1

Add $1$ and $0$ .

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

Step 2.5.2

Multiply $2^{t}$ by $1$ .

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

Step 3

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

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

Step 4

Simplify.

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Step 4.1

Apply the distributive property.

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

Step 4.2

Apply the distributive property.

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

Step 4.3

Simplify each term.

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Step 4.3.1

Multiply $- 1$ by $1$ .

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

Step 4.3.2

Rewrite $- 1 (2^{t} \ln (2))$ as $- (2^{t} \ln (2))$ .

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

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

Step 4.4

Reorder terms.

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