Calculus Examples

$\frac{d}{d v} (v^{2} (2 \sqrt{v} + 1))$

Step 1

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

$v^{2} \frac{d}{d \cdot d} [2 \sqrt{v} + 1] + (2 \sqrt{v} + 1) \frac{d}{d \cdot d} [v^{2}]$

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $2 \sqrt{v} + 1$ with respect to $d$ is $\frac{d}{d \cdot d} [2 \sqrt{v}] + \frac{d}{d \cdot d} [1]$ .

$v^{2} (\frac{d}{d \cdot d} [2 \sqrt{v}] + \frac{d}{d \cdot d} [1]) + (2 \sqrt{v} + 1) \frac{d}{d \cdot d} [v^{2}]$

Step 2.2

Since $2 \sqrt{v}$ is constant with respect to $d$ , the derivative of $2 \sqrt{v}$ with respect to $d$ is $0$ .

$v^{2} (0 + \frac{d}{d \cdot d} [1]) + (2 \sqrt{v} + 1) \frac{d}{d \cdot d} [v^{2}]$

Step 2.3

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

$v^{2} (0 + 0) + (2 \sqrt{v} + 1) \frac{d}{d \cdot d} [v^{2}]$

Step 2.4

Add $0$ and $0$ .

$v^{2} \cdot 0 + (2 \sqrt{v} + 1) \frac{d}{d \cdot d} [v^{2}]$

Step 2.5

Since $v^{2}$ is constant with respect to $d$ , the derivative of $v^{2}$ with respect to $d$ is $0$ .

$v^{2} \cdot 0 + (2 \sqrt{v} + 1) \cdot 0$

Step 3

Simplify.

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Step 3.1

Apply the distributive property.

$v^{2} \cdot 0 + 2 \sqrt{v} \cdot 0 + 1 \cdot 0$

Step 3.2

Combine terms.

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Step 3.2.1

Multiply $v^{2}$ by $0$ .

$0 + 2 \sqrt{v} \cdot 0 + 1 \cdot 0$

Step 3.2.2

Multiply $0$ by $2$ .

$0 + 0 \sqrt{v} + 1 \cdot 0$

Step 3.2.3

Multiply $0$ by $\sqrt{v}$ .

$0 + 0 + 1 \cdot 0$

Step 3.2.4

Multiply $0$ by $1$ .

$0 + 0 + 0$

Step 3.2.5

Add $0$ and $0$ .

$0 + 0$

Step 3.2.6

Add $0$ and $0$ .

$0$