Find the Derivative - d/dt tan( square root of 5t)

$\tan (\sqrt{5 t})$

Step 1

Simplify with factoring out.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5 t}$ as ${(5 t)}^{\frac{1}{2}}$ .

$\frac{d}{d t} [\tan ({(5 t)}^{\frac{1}{2}})]$

Factor $5$ out of $5 t$ .

$\frac{d}{d t} [\tan ({(5 (t))}^{\frac{1}{2}})]$

Apply the product rule to $5 (t)$ .

$\frac{d}{d t} [\tan (5^{\frac{1}{2}} t^{\frac{1}{2}})]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \tan (t)$ and $g (t) = 5^{\frac{1}{2}} t^{\frac{1}{2}}$ .

To apply the Chain Rule, set $u$ as $5^{\frac{1}{2}} t^{\frac{1}{2}}$ .

$\frac{d}{d u} [\tan (u)] \frac{d}{d t} [5^{\frac{1}{2}} t^{\frac{1}{2}}]$

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

$\sec^{2} (u) \frac{d}{d t} [5^{\frac{1}{2}} t^{\frac{1}{2}}]$

Replace all occurrences of $u$ with $5^{\frac{1}{2}} t^{\frac{1}{2}}$ .

$\sec^{2} (5^{\frac{1}{2}} t^{\frac{1}{2}}) \frac{d}{d t} [5^{\frac{1}{2}} t^{\frac{1}{2}}]$

Step 3

Differentiate.

Since $5^{\frac{1}{2}}$ is constant with respect to $t$ , the derivative of $5^{\frac{1}{2}} t^{\frac{1}{2}}$ with respect to $t$ is $5^{\frac{1}{2}} \frac{d}{d t} [t^{\frac{1}{2}}]$ .

$\sec^{2} (5^{\frac{1}{2}} t^{\frac{1}{2}}) (5^{\frac{1}{2}} \frac{d}{d t} [t^{\frac{1}{2}}])$

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

$\sec^{2} (5^{\frac{1}{2}} t^{\frac{1}{2}}) (5^{\frac{1}{2}} (\frac{1}{2} t^{\frac{1}{2} - 1}))$

Step 4

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

$\sec^{2} (5^{\frac{1}{2}} t^{\frac{1}{2}}) (5^{\frac{1}{2}} (\frac{1}{2} t^{\frac{1}{2} - 1 \cdot \frac{2}{2}}))$

Step 5

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

$\sec^{2} (5^{\frac{1}{2}} t^{\frac{1}{2}}) (5^{\frac{1}{2}} (\frac{1}{2} t^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}}))$

Step 6

Combine the numerators over the common denominator.

$\sec^{2} (5^{\frac{1}{2}} t^{\frac{1}{2}}) (5^{\frac{1}{2}} (\frac{1}{2} t^{\frac{1 - 1 \cdot 2}{2}}))$

Step 7

Simplify the numerator.

Multiply $- 1$ by $2$ .

$\sec^{2} (5^{\frac{1}{2}} t^{\frac{1}{2}}) (5^{\frac{1}{2}} (\frac{1}{2} t^{\frac{1 - 2}{2}}))$

Subtract $2$ from $1$ .

$\sec^{2} (5^{\frac{1}{2}} t^{\frac{1}{2}}) (5^{\frac{1}{2}} (\frac{1}{2} t^{\frac{- 1}{2}}))$

Step 8

Move the negative in front of the fraction.

$\sec^{2} (5^{\frac{1}{2}} t^{\frac{1}{2}}) (5^{\frac{1}{2}} (\frac{1}{2} t^{- \frac{1}{2}}))$

Step 9

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

$\sec^{2} (5^{\frac{1}{2}} t^{\frac{1}{2}}) (5^{\frac{1}{2}} \frac{t^{- \frac{1}{2}}}{2})$

Step 10

Combine $5^{\frac{1}{2}}$ and $\frac{t^{- \frac{1}{2}}}{2}$ .

$\sec^{2} (5^{\frac{1}{2}} t^{\frac{1}{2}}) \frac{5^{\frac{1}{2}} t^{- \frac{1}{2}}}{2}$

Step 11

Move $t^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\sec^{2} (5^{\frac{1}{2}} t^{\frac{1}{2}}) \frac{5^{\frac{1}{2}}}{2 t^{\frac{1}{2}}}$

Step 12

Combine $\sec^{2} (5^{\frac{1}{2}} t^{\frac{1}{2}})$ and $\frac{5^{\frac{1}{2}}}{2 t^{\frac{1}{2}}}$ .

$\frac{\sec^{2} (5^{\frac{1}{2}} t^{\frac{1}{2}}) \cdot 5^{\frac{1}{2}}}{2 t^{\frac{1}{2}}}$

Step 13

Reorder factors in $\sec^{2} (5^{\frac{1}{2}} t^{\frac{1}{2}}) \cdot 5^{\frac{1}{2}}$ .

$\frac{5^{\frac{1}{2}} \sec^{2} (5^{\frac{1}{2}} t^{\frac{1}{2}})}{2 t^{\frac{1}{2}}}$

Find the Derivative - d/dt tan( square root of 5t)

Related Questions

Support LCMGCF.com by disabling your adblocker.