Calculus Examples

$2 t^{2} + \sqrt{t^{3}}$

Step 1

By the Sum Rule, the derivative of $2 t^{2} + \sqrt{t^{3}}$ with respect to $t$ is $\frac{d}{d t} [2 t^{2}] + \frac{d}{d t} [\sqrt{t^{3}}]$ .

$\frac{d}{d t} [2 t^{2}] + \frac{d}{d t} [\sqrt{t^{3}}]$

Step 2

Evaluate $\frac{d}{d t} [2 t^{2}]$ .

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Step 2.1

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

$2 \frac{d}{d t} [t^{2}] + \frac{d}{d t} [\sqrt{t^{3}}]$

Step 2.2

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

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

Step 2.3

Multiply $2$ by $2$ .

$4 t + \frac{d}{d t} [\sqrt{t^{3}}]$

Step 3

Evaluate $\frac{d}{d t} [\sqrt{t^{3}}]$ .

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Step 3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{t^{3}}$ as $t^{\frac{3}{2}}$ .

$4 t + \frac{d}{d t} [t^{\frac{3}{2}}]$

Step 3.2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = \frac{3}{2}$ .

$4 t + \frac{3}{2} t^{\frac{3}{2} - 1}$

Step 3.3

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

$4 t + \frac{3}{2} t^{\frac{3}{2} - 1 \cdot \frac{2}{2}}$

Step 3.4

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

$4 t + \frac{3}{2} t^{\frac{3}{2} + \frac{- 1 \cdot 2}{2}}$

Step 3.5

Combine the numerators over the common denominator.

$4 t + \frac{3}{2} t^{\frac{3 - 1 \cdot 2}{2}}$

Step 3.6

Simplify the numerator.

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Step 3.6.1

Multiply $- 1$ by $2$ .

$4 t + \frac{3}{2} t^{\frac{3 - 2}{2}}$

Step 3.6.2

Subtract $2$ from $3$ .

$4 t + \frac{3}{2} t^{\frac{1}{2}}$

Step 4

Simplify.

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Step 4.1

Reorder terms.

$\frac{3}{2} t^{\frac{1}{2}} + 4 t$

Step 4.2

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

$\frac{3 t^{\frac{1}{2}}}{2} + 4 t$