Calculus Examples

$\int_{\tan (x)}^{x^{2}} \frac{1}{\sqrt{2 + t^{4}}} d t$

Multiply $\frac{1}{\sqrt{2 + t^{4}}}$ by $\frac{\sqrt{2 + t^{4}}}{\sqrt{2 + t^{4}}}$ .

$\frac{d}{d x} [\int_{\tan (x)}^{x^{2}} \frac{1}{\sqrt{2 + t^{4}}} \cdot \frac{\sqrt{2 + t^{4}}}{\sqrt{2 + t^{4}}} d t]$

Combine and simplify the denominator.

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Multiply $\frac{1}{\sqrt{2 + t^{4}}}$ by $\frac{\sqrt{2 + t^{4}}}{\sqrt{2 + t^{4}}}$ .

$\frac{d}{d x} [\int_{\tan (x)}^{x^{2}} \frac{\sqrt{2 + t^{4}}}{\sqrt{2 + t^{4}} \sqrt{2 + t^{4}}} d t]$

Raise $\sqrt{2 + t^{4}}$ to the power of $1$ .

$\frac{d}{d x} [\int_{\tan (x)}^{x^{2}} \frac{\sqrt{2 + t^{4}}}{{\sqrt{2 + t^{4}}}^{1} \sqrt{2 + t^{4}}} d t]$

Raise $\sqrt{2 + t^{4}}$ to the power of $1$ .

$\frac{d}{d x} [\int_{\tan (x)}^{x^{2}} \frac{\sqrt{2 + t^{4}}}{{\sqrt{2 + t^{4}}}^{1} {\sqrt{2 + t^{4}}}^{1}} d t]$

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

$\frac{d}{d x} [\int_{\tan (x)}^{x^{2}} \frac{\sqrt{2 + t^{4}}}{{\sqrt{2 + t^{4}}}^{1 + 1}} d t]$

Add $1$ and $1$ .

$\frac{d}{d x} [\int_{\tan (x)}^{x^{2}} \frac{\sqrt{2 + t^{4}}}{{\sqrt{2 + t^{4}}}^{2}} d t]$

Rewrite ${\sqrt{2 + t^{4}}}^{2}$ as $2 + t^{4}$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 + t^{4}}$ as ${(2 + t^{4})}^{\frac{1}{2}}$ .

$\frac{d}{d x} [\int_{\tan (x)}^{x^{2}} \frac{\sqrt{2 + t^{4}}}{{({(2 + t^{4})}^{\frac{1}{2}})}^{2}} d t]$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{d}{d x} [\int_{\tan (x)}^{x^{2}} \frac{\sqrt{2 + t^{4}}}{{(2 + t^{4})}^{\frac{1}{2} \cdot 2}} d t]$

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

$\frac{d}{d x} [\int_{\tan (x)}^{x^{2}} \frac{\sqrt{2 + t^{4}}}{{(2 + t^{4})}^{\frac{2}{2}}} d t]$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{d}{d x} [\int_{\tan (x)}^{x^{2}} \frac{\sqrt{2 + t^{4}}}{{(2 + t^{4})}^{\frac{2}{2}}} d t]$

Rewrite the expression.

$\frac{d}{d x} [\int_{\tan (x)}^{x^{2}} \frac{\sqrt{2 + t^{4}}}{{(2 + t^{4})}^{1}} d t]$

Simplify.

$\frac{d}{d x} [\int_{\tan (x)}^{x^{2}} \frac{\sqrt{2 + t^{4}}}{2 + t^{4}} d t]$

Split the integral into two integrals where $c$ is some value between $\tan (x)$ and $x^{2}$ .

$\frac{d}{d x} [\int_{\tan (x)}^{c} \frac{\sqrt{2 + t^{4}}}{2 + t^{4}} d t + \int_{c}^{x^{2}} \frac{\sqrt{2 + t^{4}}}{2 + t^{4}} d t]$

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

$\frac{d}{d x} [\int_{\tan (x)}^{c} \frac{\sqrt{2 + t^{4}}}{2 + t^{4}} d t] + \frac{d}{d x} [\int_{c}^{x^{2}} \frac{\sqrt{2 + t^{4}}}{2 + t^{4}} d t]$

Swap the bounds of integration.

$\frac{d}{d x} [- \int_{c}^{\tan (x)} \frac{\sqrt{2 + t^{4}}}{2 + t^{4}} d t] + \frac{d}{d x} [\int_{c}^{x^{2}} \frac{\sqrt{2 + t^{4}}}{2 + t^{4}} d t]$

Take the derivative of $- \int_{c}^{\tan (x)} \frac{\sqrt{2 + t^{4}}}{2 + t^{4}} d t$ with respect to $x$ using Fundamental Theorem of Calculus and the chain rule.

$\frac{d}{d x} [\tan (x)] (- \frac{\sqrt{2 + \tan^{4} (x)}}{2 + \tan^{4} (x)}) + \frac{d}{d x} [\int_{c}^{x^{2}} \frac{\sqrt{2 + t^{4}}}{2 + t^{4}} d t]$

The derivative of $\tan (x)$ with respect to $x$ is $\sec^{2} (x)$ .

$\sec^{2} (x) (- \frac{\sqrt{2 + \tan^{4} (x)}}{2 + \tan^{4} (x)}) + \frac{d}{d x} [\int_{c}^{x^{2}} \frac{\sqrt{2 + t^{4}}}{2 + t^{4}} d t]$

Take the derivative of $\int_{c}^{x^{2}} \frac{\sqrt{2 + t^{4}}}{2 + t^{4}} d t$ with respect to $x$ using Fundamental Theorem of Calculus and the chain rule.

$\sec^{2} (x) (- \frac{\sqrt{2 + \tan^{4} (x)}}{2 + \tan^{4} (x)}) + \frac{d}{d x} [x^{2}] \frac{\sqrt{2 + {(x^{2})}^{4}}}{2 + {(x^{2})}^{4}}$

Differentiate using the Power Rule.

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

$\sec^{2} (x) (- \frac{\sqrt{2 + \tan^{4} (x)}}{2 + \tan^{4} (x)}) + 2 x \frac{\sqrt{2 + {(x^{2})}^{4}}}{2 + {(x^{2})}^{4}}$

Simplify terms.

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Combine $\sec^{2} (x)$ and $\frac{\sqrt{2 + \tan^{4} (x)}}{2 + \tan^{4} (x)}$ .

$- \frac{\sec^{2} (x) \sqrt{2 + \tan^{4} (x)}}{2 + \tan^{4} (x)} + 2 x \frac{\sqrt{2 + {(x^{2})}^{4}}}{2 + {(x^{2})}^{4}}$

Combine fractions.

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Apply basic rules of exponents.

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Multiply the exponents in ${(x^{2})}^{4}$ .

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Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$- \frac{\sec^{2} (x) \sqrt{2 + \tan^{4} (x)}}{2 + \tan^{4} (x)} + 2 x \frac{\sqrt{2 + x^{2 \cdot 4}}}{2 + {(x^{2})}^{4}}$

Multiply $2$ by $4$ .

$- \frac{\sec^{2} (x) \sqrt{2 + \tan^{4} (x)}}{2 + \tan^{4} (x)} + 2 x \frac{\sqrt{2 + x^{8}}}{2 + {(x^{2})}^{4}}$

Multiply the exponents in ${(x^{2})}^{4}$ .

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Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$- \frac{\sec^{2} (x) \sqrt{2 + \tan^{4} (x)}}{2 + \tan^{4} (x)} + 2 x \frac{\sqrt{2 + x^{8}}}{2 + x^{2 \cdot 4}}$

Multiply $2$ by $4$ .

$- \frac{\sec^{2} (x) \sqrt{2 + \tan^{4} (x)}}{2 + \tan^{4} (x)} + 2 x \frac{\sqrt{2 + x^{8}}}{2 + x^{8}}$

Combine $2$ and $\frac{\sqrt{2 + x^{8}}}{2 + x^{8}}$ .

$- \frac{\sec^{2} (x) \sqrt{2 + \tan^{4} (x)}}{2 + \tan^{4} (x)} + x \frac{2 \sqrt{2 + x^{8}}}{2 + x^{8}}$

Combine $x$ and $\frac{2 \sqrt{2 + x^{8}}}{2 + x^{8}}$ .

$- \frac{\sec^{2} (x) \sqrt{2 + \tan^{4} (x)}}{2 + \tan^{4} (x)} + \frac{x (2 \sqrt{2 + x^{8}})}{2 + x^{8}}$

Move $2$ to the left of $x$ .

$- \frac{\sec^{2} (x) \sqrt{2 + \tan^{4} (x)}}{2 + \tan^{4} (x)} + \frac{2 x \sqrt{2 + x^{8}}}{2 + x^{8}}$