Calculus Examples

$y = e^{p} (p + p \sqrt{p})$

Step 1

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

$\frac{d}{d p} [e^{p} (p + p \cdot p^{\frac{1}{2}})]$

Step 2

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

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Step 2.1

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

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Step 2.1.1

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

$\frac{d}{d p} [e^{p} (p + p^{1} p^{\frac{1}{2}})]$

Step 2.1.2

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

$\frac{d}{d p} [e^{p} (p + p^{1 + \frac{1}{2}})]$

Step 2.2

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

$\frac{d}{d p} [e^{p} (p + p^{\frac{2}{2} + \frac{1}{2}})]$

Step 2.3

Combine the numerators over the common denominator.

$\frac{d}{d p} [e^{p} (p + p^{\frac{2 + 1}{2}})]$

Step 2.4

Add $2$ and $1$ .

$\frac{d}{d p} [e^{p} (p + p^{\frac{3}{2}})]$

Step 3

Differentiate using the Product Rule which states that $\frac{d}{d p} [f (p) g (p)]$ is $f (p) \frac{d}{d p} [g (p)] + g (p) \frac{d}{d p} [f (p)]$ where $f (p) = e^{p}$ and $g (p) = p + p^{\frac{3}{2}}$ .

$e^{p} \frac{d}{d p} [p + p^{\frac{3}{2}}] + (p + p^{\frac{3}{2}}) \frac{d}{d p} [e^{p}]$

Step 4

Differentiate.

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Step 4.1

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

$e^{p} (\frac{d}{d p} [p] + \frac{d}{d p} [p^{\frac{3}{2}}]) + (p + p^{\frac{3}{2}}) \frac{d}{d p} [e^{p}]$

Step 4.2

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

$e^{p} (1 + \frac{d}{d p} [p^{\frac{3}{2}}]) + (p + p^{\frac{3}{2}}) \frac{d}{d p} [e^{p}]$

Step 4.3

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

$e^{p} (1 + \frac{3}{2} p^{\frac{3}{2} - 1}) + (p + p^{\frac{3}{2}}) \frac{d}{d p} [e^{p}]$

Step 5

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

$e^{p} (1 + \frac{3}{2} p^{\frac{3}{2} - 1 \cdot \frac{2}{2}}) + (p + p^{\frac{3}{2}}) \frac{d}{d p} [e^{p}]$

Step 6

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

$e^{p} (1 + \frac{3}{2} p^{\frac{3}{2} + \frac{- 1 \cdot 2}{2}}) + (p + p^{\frac{3}{2}}) \frac{d}{d p} [e^{p}]$

Step 7

Combine the numerators over the common denominator.

$e^{p} (1 + \frac{3}{2} p^{\frac{3 - 1 \cdot 2}{2}}) + (p + p^{\frac{3}{2}}) \frac{d}{d p} [e^{p}]$

Step 8

Simplify the numerator.

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Step 8.1

Multiply $- 1$ by $2$ .

$e^{p} (1 + \frac{3}{2} p^{\frac{3 - 2}{2}}) + (p + p^{\frac{3}{2}}) \frac{d}{d p} [e^{p}]$

Step 8.2

Subtract $2$ from $3$ .

$e^{p} (1 + \frac{3}{2} p^{\frac{1}{2}}) + (p + p^{\frac{3}{2}}) \frac{d}{d p} [e^{p}]$

Step 9

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

$e^{p} (1 + \frac{3 p^{\frac{1}{2}}}{2}) + (p + p^{\frac{3}{2}}) \frac{d}{d p} [e^{p}]$

Step 10

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

$e^{p} (1 + \frac{3 p^{\frac{1}{2}}}{2}) + (p + p^{\frac{3}{2}}) e^{p}$

Step 11

Simplify.

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Step 11.1

Apply the distributive property.

$e^{p} \cdot 1 + e^{p} \frac{3 p^{\frac{1}{2}}}{2} + (p + p^{\frac{3}{2}}) e^{p}$

Step 11.2

Apply the distributive property.

$e^{p} \cdot 1 + e^{p} \frac{3 p^{\frac{1}{2}}}{2} + p e^{p} + p^{\frac{3}{2}} e^{p}$

Step 11.3

Combine terms.

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Step 11.3.1

Multiply $e^{p}$ by $1$ .

$e^{p} + e^{p} \frac{3 p^{\frac{1}{2}}}{2} + p e^{p} + p^{\frac{3}{2}} e^{p}$

Step 11.3.2

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

$e^{p} + \frac{e^{p} (3 p^{\frac{1}{2}})}{2} + p e^{p} + p^{\frac{3}{2}} e^{p}$

Step 11.4

Reorder terms.

$e^{p} p + e^{p} p^{\frac{3}{2}} + e^{p} + \frac{e^{p} (3 p^{\frac{1}{2}})}{2}$