Calculus Examples

$\int_{2}^{1} 6 \sqrt{x} - \frac{1}{\sqrt{2 x}} d x$

Step 1

Simplify with factoring out.

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Step 1.1

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

$\frac{d}{d x} [\int_{2}^{1} 6 x^{\frac{1}{2}} - \frac{1}{\sqrt{2 x}} d x]$

Step 1.2

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

$\frac{d}{d x} [\int_{2}^{1} 6 x^{\frac{1}{2}} - \frac{1}{{(2 x)}^{\frac{1}{2}}} d x]$

Step 1.3

Factor $2$ out of $2 x$ .

$\frac{d}{d x} [\int_{2}^{1} 6 x^{\frac{1}{2}} - \frac{1}{{(2 (x))}^{\frac{1}{2}}} d x]$

Step 1.4

Apply the product rule to $2 (x)$ .

$\frac{d}{d x} [\int_{2}^{1} 6 x^{\frac{1}{2}} - \frac{1}{2^{\frac{1}{2}} x^{\frac{1}{2}}} d x]$

Step 2

Once $\int_{2}^{1} 6 x^{\frac{1}{2}} - \frac{1}{2^{\frac{1}{2}} x^{\frac{1}{2}}} d x$ has been evaluated, it will be constant with respect to $x$ , so the derivative of $\int_{2}^{1} 6 x^{\frac{1}{2}} - \frac{1}{2^{\frac{1}{2}} x^{\frac{1}{2}}} d x$ with respect to $x$ is $0$ .

$0$