Calculus Examples

$u = \sqrt[3]{t} + 6 \sqrt{t^{3}}$

Step 1

Factor out $t^{2}$ .

$\frac{d}{d u} [\sqrt[3]{t} + 6 \sqrt{t^{2} t}]$

Step 2

Differentiate.

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Step 2.1

Pull terms out from under the radical.

$\frac{d}{d u} [\sqrt[3]{t} + 6 (t \sqrt{t})]$

Step 2.2

By the Sum Rule, the derivative of $\sqrt[3]{t} + 6 (t \sqrt{t})$ with respect to $u$ is $\frac{d}{d u} [\sqrt[3]{t}] + \frac{d}{d u} [6 (t \sqrt{t})]$ .

$\frac{d}{d u} [\sqrt[3]{t}] + \frac{d}{d u} [6 (t \sqrt{t})]$

Step 2.3

Since $\sqrt[3]{t}$ is constant with respect to $u$ , the derivative of $\sqrt[3]{t}$ with respect to $u$ is $0$ .

$0 + \frac{d}{d u} [6 (t \sqrt{t})]$

Step 3

Evaluate $\frac{d}{d u} [6 (t \sqrt{t})]$ .

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Step 3.1

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

$0 + \frac{d}{d u} [6 (t \cdot t^{\frac{1}{2}})]$

Step 3.2

Multiply $t$ by $t^{\frac{1}{2}}$ by adding the exponents.

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Step 3.2.1

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

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Step 3.2.1.1

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

$0 + \frac{d}{d u} [6 (t^{1} t^{\frac{1}{2}})]$

Step 3.2.1.2

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

$0 + \frac{d}{d u} [6 t^{1 + \frac{1}{2}}]$

Step 3.2.2

Write $1$ as a fraction with a common denominator.

$0 + \frac{d}{d u} [6 t^{\frac{2}{2} + \frac{1}{2}}]$

Step 3.2.3

Combine the numerators over the common denominator.

$0 + \frac{d}{d u} [6 t^{\frac{2 + 1}{2}}]$

Step 3.2.4

Add $2$ and $1$ .

$0 + \frac{d}{d u} [6 t^{\frac{3}{2}}]$

Step 3.3

Since $6 t^{\frac{3}{2}}$ is constant with respect to $u$ , the derivative of $6 t^{\frac{3}{2}}$ with respect to $u$ is $0$ .

$0 + 0$

Step 4

Add $0$ and $0$ .

$0$