Calculus Examples

$10 t^{2} \cdot 2^{- t}$

Step 1

Since $10$ is constant with respect to $t$ , the derivative of $10 t^{2} \cdot 2^{- t}$ with respect to $t$ is $10 \frac{d}{d t} [t^{2} \cdot 2^{- t}]$ .

$10 \frac{d}{d t} [t^{2} \cdot 2^{- t}]$

Step 2

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

$10 (t^{2} \frac{d}{d t} [2^{- t}] + 2^{- t} \frac{d}{d t} [t^{2}])$

Step 3

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 $g (t) = - t$ .

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Step 3.1

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

$10 (t^{2} (\frac{d}{d u} [2^{u}] \frac{d}{d t} [- t]) + 2^{- t} \frac{d}{d t} [t^{2}])$

Step 3.2

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

$10 (t^{2} (2^{u} \ln (2) \frac{d}{d t} [- t]) + 2^{- t} \frac{d}{d t} [t^{2}])$

Step 3.3

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

$10 (t^{2} (2^{- t} \ln (2) \frac{d}{d t} [- t]) + 2^{- t} \frac{d}{d t} [t^{2}])$

Step 4

Differentiate.

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Step 4.1

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

$10 (t^{2} (2^{- t} \ln (2) (- \frac{d}{d t} [t])) + 2^{- t} \frac{d}{d t} [t^{2}])$

Step 4.2

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

$10 (t^{2} (2^{- t} \ln (2) (- 1 \cdot 1)) + 2^{- t} \frac{d}{d t} [t^{2}])$

Step 4.3

Simplify the expression.

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Step 4.3.1

Multiply $- 1$ by $1$ .

$10 (t^{2} (2^{- t} \ln (2) \cdot - 1) + 2^{- t} \frac{d}{d t} [t^{2}])$

Step 4.3.2

Move $- 1$ to the left of $2^{- t} \ln (2)$ .

$10 (t^{2} (- 1 \cdot (2^{- t} \ln (2))) + 2^{- t} \frac{d}{d t} [t^{2}])$

Step 4.3.3

Rewrite $- 1 \cdot 2^{- t}$ as $- 2^{- t}$ .

$10 (t^{2} (- 2^{- t} \ln (2)) + 2^{- t} \frac{d}{d t} [t^{2}])$

Step 4.4

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

$10 (t^{2} (- 2^{- t} \ln (2)) + 2^{- t} (2 t))$

Step 5

Multiply $2^{- t}$ by $2$ by adding the exponents.

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Step 5.1

Move $2$ .

$10 (t^{2} (- 2^{- t} \ln (2)) + 2 \cdot 2^{- t} t)$

Step 5.2

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

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Step 5.2.1

Raise $2$ to the power of $1$ .

$10 (t^{2} (- 2^{- t} \ln (2)) + 2^{1} \cdot 2^{- t} t)$

Step 5.2.2

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

$10 (t^{2} (- 2^{- t} \ln (2)) + 2^{1 - t} t)$

Step 6

Simplify.

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Step 6.1

Apply the distributive property.

$10 (t^{2} (- 2^{- t} \ln (2))) + 10 (2^{1 - t} t)$

Step 6.2

Multiply $- 1$ by $10$ .

$- 10 t^{2} (2^{- t} \ln (2)) + 10 \cdot 2^{1 - t} t$

Step 6.3

Reorder terms.

$- 10 \cdot 2^{- t} t^{2} \ln (2) + 10 \cdot 2^{1 - t} t$

Step 6.4

Reorder factors in $- 10 \cdot 2^{- t} t^{2} \ln (2) + 10 \cdot 2^{1 - t} t$ .

$- 10 t^{2} \cdot 2^{- t} \ln (2) + 10 t \cdot 2^{1 - t}$