Calculus Examples

$q = \tan (\frac{t}{\sqrt{t + 8}})$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d q} [f (g (q))]$ is $f' (g (q)) g' (q)$ where $f (q) = \tan (q)$ and $g (q) = \frac{t}{\sqrt{t + 8}}$ .

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Step 1.1

To apply the Chain Rule, set $u$ as $\frac{t}{\sqrt{t + 8}}$ .

$\frac{d}{d u} [\tan (u)] \frac{d}{d q} [\frac{t}{\sqrt{t + 8}}]$

Step 1.2

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

$\sec^{2} (u) \frac{d}{d q} [\frac{t}{\sqrt{t + 8}}]$

Step 1.3

Replace all occurrences of $u$ with $\frac{t}{\sqrt{t + 8}}$ .

$\sec^{2} (\frac{t}{\sqrt{t + 8}}) \frac{d}{d q} [\frac{t}{\sqrt{t + 8}}]$

Step 2

Multiply $\frac{t}{\sqrt{t + 8}}$ by $\frac{\sqrt{t + 8}}{\sqrt{t + 8}}$ .

$\sec^{2} (\frac{t}{\sqrt{t + 8}}) \frac{d}{d q} [\frac{t}{\sqrt{t + 8}} \cdot \frac{\sqrt{t + 8}}{\sqrt{t + 8}}]$

Step 3

Combine and simplify the denominator.

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Step 3.1

Multiply $\frac{t}{\sqrt{t + 8}}$ by $\frac{\sqrt{t + 8}}{\sqrt{t + 8}}$ .

$\sec^{2} (\frac{t}{\sqrt{t + 8}}) \frac{d}{d q} [\frac{t \sqrt{t + 8}}{\sqrt{t + 8} \sqrt{t + 8}}]$

Step 3.2

Raise $\sqrt{t + 8}$ to the power of $1$ .

$\sec^{2} (\frac{t}{\sqrt{t + 8}}) \frac{d}{d q} [\frac{t \sqrt{t + 8}}{{\sqrt{t + 8}}^{1} \sqrt{t + 8}}]$

Step 3.3

Raise $\sqrt{t + 8}$ to the power of $1$ .

$\sec^{2} (\frac{t}{\sqrt{t + 8}}) \frac{d}{d q} [\frac{t \sqrt{t + 8}}{{\sqrt{t + 8}}^{1} {\sqrt{t + 8}}^{1}}]$

Step 3.4

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

$\sec^{2} (\frac{t}{\sqrt{t + 8}}) \frac{d}{d q} [\frac{t \sqrt{t + 8}}{{\sqrt{t + 8}}^{1 + 1}}]$

Step 3.5

Add $1$ and $1$ .

$\sec^{2} (\frac{t}{\sqrt{t + 8}}) \frac{d}{d q} [\frac{t \sqrt{t + 8}}{{\sqrt{t + 8}}^{2}}]$

Step 3.6

Rewrite ${\sqrt{t + 8}}^{2}$ as $t + 8$ .

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Step 3.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{t + 8}$ as ${(t + 8)}^{\frac{1}{2}}$ .

$\sec^{2} (\frac{t}{\sqrt{t + 8}}) \frac{d}{d q} [\frac{t \sqrt{t + 8}}{{({(t + 8)}^{\frac{1}{2}})}^{2}}]$

Step 3.6.2

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

$\sec^{2} (\frac{t}{\sqrt{t + 8}}) \frac{d}{d q} [\frac{t \sqrt{t + 8}}{{(t + 8)}^{\frac{1}{2} \cdot 2}}]$

Step 3.6.3

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

$\sec^{2} (\frac{t}{\sqrt{t + 8}}) \frac{d}{d q} [\frac{t \sqrt{t + 8}}{{(t + 8)}^{\frac{2}{2}}}]$

Step 3.6.4

Cancel the common factor of $2$ .

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Step 3.6.4.1

Cancel the common factor.

$\sec^{2} (\frac{t}{\sqrt{t + 8}}) \frac{d}{d q} [\frac{t \sqrt{t + 8}}{{(t + 8)}^{\frac{2}{2}}}]$

Step 3.6.4.2

Rewrite the expression.

$\sec^{2} (\frac{t}{\sqrt{t + 8}}) \frac{d}{d q} [\frac{t \sqrt{t + 8}}{{(t + 8)}^{1}}]$

Step 3.6.5

Simplify.

$\sec^{2} (\frac{t}{\sqrt{t + 8}}) \frac{d}{d q} [\frac{t \sqrt{t + 8}}{t + 8}]$

Step 4

Since $\frac{t \sqrt{t + 8}}{t + 8}$ is constant with respect to $q$ , the derivative of $\frac{t \sqrt{t + 8}}{t + 8}$ with respect to $q$ is $0$ .

$\sec^{2} (\frac{t}{\sqrt{t + 8}}) \cdot 0$

Step 5

Multiply $\sec^{2} (\frac{t}{\sqrt{t + 8}})$ by $0$ .

$0$