Calculus Examples

$u = \sqrt[5]{t} + 4 \sqrt{t^{5}}$

Step 1

Rewrite $t^{5}$ as ${(t^{2})}^{2} t$ .

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Step 1.1

Factor out $t^{4}$ .

$\frac{d}{d u} [\sqrt[5]{t} + 4 \sqrt{t^{4} t}]$

Step 1.2

Rewrite $t^{4}$ as ${(t^{2})}^{2}$ .

$\frac{d}{d u} [\sqrt[5]{t} + 4 \sqrt{{(t^{2})}^{2} t}]$

Step 2

Differentiate.

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Step 2.1

Pull terms out from under the radical.

$\frac{d}{d u} [\sqrt[5]{t} + 4 (t^{2} \sqrt{t})]$

Step 2.2

By the Sum Rule, the derivative of $\sqrt[5]{t} + 4 (t^{2} \sqrt{t})$ with respect to $u$ is $\frac{d}{d u} [\sqrt[5]{t}] + \frac{d}{d u} [4 (t^{2} \sqrt{t})]$ .

$\frac{d}{d u} [\sqrt[5]{t}] + \frac{d}{d u} [4 (t^{2} \sqrt{t})]$

Step 2.3

Since $\sqrt[5]{t}$ is constant with respect to $u$ , the derivative of $\sqrt[5]{t}$ with respect to $u$ is $0$ .

$0 + \frac{d}{d u} [4 (t^{2} \sqrt{t})]$

Step 3

Evaluate $\frac{d}{d u} [4 (t^{2} \sqrt{t})]$ .

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Step 3.1

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

$0 + \frac{d}{d u} [4 (t^{2} t^{\frac{1}{2}})]$

Step 3.2

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

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Step 3.2.1

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

$0 + \frac{d}{d u} [4 t^{2 + \frac{1}{2}}]$

Step 3.2.2

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

$0 + \frac{d}{d u} [4 t^{2 \cdot \frac{2}{2} + \frac{1}{2}}]$

Step 3.2.3

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

$0 + \frac{d}{d u} [4 t^{\frac{2 \cdot 2}{2} + \frac{1}{2}}]$

Step 3.2.4

Combine the numerators over the common denominator.

$0 + \frac{d}{d u} [4 t^{\frac{2 \cdot 2 + 1}{2}}]$

Step 3.2.5

Simplify the numerator.

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Step 3.2.5.1

Multiply $2$ by $2$ .

$0 + \frac{d}{d u} [4 t^{\frac{4 + 1}{2}}]$

Step 3.2.5.2

Add $4$ and $1$ .

$0 + \frac{d}{d u} [4 t^{\frac{5}{2}}]$

Step 3.3

Since $4 t^{\frac{5}{2}}$ is constant with respect to $u$ , the derivative of $4 t^{\frac{5}{2}}$ with respect to $u$ is $0$ .

$0 + 0$

Step 4

Add $0$ and $0$ .

$0$