Calculus Examples

$\int_{p}^{- p} \sqrt{\sqrt[3]{x}} d x$

Step 1

Rewrite $\sqrt{\sqrt[3]{x}}$ as $\sqrt[6]{x}$ .

$\frac{d}{d p} [\int_{p}^{- p} \sqrt[6]{x} d x]$

Step 2

Split the integral into two integrals where $c$ is some value between $p$ and $- p$ .

$\frac{d}{d p} [\int_{p}^{c} \sqrt[6]{x} d x + \int_{c}^{- p} \sqrt[6]{x} d x]$

Step 3

By the Sum Rule, the derivative of $\int_{p}^{c} \sqrt[6]{x} d x + \int_{c}^{- p} \sqrt[6]{x} d x$ with respect to $p$ is $\frac{d}{d p} [\int_{p}^{c} \sqrt[6]{x} d x] + \frac{d}{d p} [\int_{c}^{- p} \sqrt[6]{x} d x]$ .

$\frac{d}{d p} [\int_{p}^{c} \sqrt[6]{x} d x] + \frac{d}{d p} [\int_{c}^{- p} \sqrt[6]{x} d x]$

Step 4

Swap the bounds of integration.

$\frac{d}{d p} [- \int_{c}^{p} \sqrt[6]{x} d x] + \frac{d}{d p} [\int_{c}^{- p} \sqrt[6]{x} d x]$

Step 5

Take the derivative of $- \int_{c}^{p} \sqrt[6]{x} d x$ with respect to $p$ using Fundamental Theorem of Calculus.

$- \sqrt[6]{p} + \frac{d}{d p} [\int_{c}^{- p} \sqrt[6]{x} d x]$

Step 6

Take the derivative of $\int_{c}^{- p} \sqrt[6]{x} d x$ with respect to $p$ using Fundamental Theorem of Calculus and the chain rule.

$- \sqrt[6]{p} + \frac{d}{d p} [- p] \sqrt[6]{- p}$

Step 7

Differentiate.

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Step 7.1

Since $- 1$ is constant with respect to $p$ , the derivative of $- p$ with respect to $p$ is $- \frac{d}{d p} [p]$ .

$- \sqrt[6]{p} - \frac{d}{d p} [p] \sqrt[6]{- p}$

Step 7.2

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

$- \sqrt[6]{p} - 1 \cdot 1 \sqrt[6]{- p}$

Step 7.3

Multiply $- 1$ by $1$ .

$- \sqrt[6]{p} - \sqrt[6]{- p}$