Find the Derivative – d/dt y=tan( square root of 3t)

$y = \tan (\sqrt{3 t})$

Simplify with factoring out.

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Rewrite $\sqrt{3 t}$ as ${(3 t)}^{\frac{1}{2}}$ .

$\frac{d}{d t} [\tan ({(3 t)}^{\frac{1}{2}})]$

Factor $3$ out of $3 t$ .

$\frac{d}{d t} [\tan ({(3 (t))}^{\frac{1}{2}})]$

Apply the product rule to $3 (t)$ .

$\frac{d}{d t} [\tan (3^{\frac{1}{2}} t^{\frac{1}{2}})]$

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 $g (t) = 3^{\frac{1}{2}} t^{\frac{1}{2}}$ .

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To apply the Chain Rule, set $u$ as $3^{\frac{1}{2}} t^{\frac{1}{2}}$ .

$\frac{d}{d u} [\tan (u)] \frac{d}{d t} [3^{\frac{1}{2}} t^{\frac{1}{2}}]$

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

$\sec^{2} (u) \frac{d}{d t} [3^{\frac{1}{2}} t^{\frac{1}{2}}]$

Replace all occurrences of $u$ with $3^{\frac{1}{2}} t^{\frac{1}{2}}$ .

$\sec^{2} (3^{\frac{1}{2}} t^{\frac{1}{2}}) \frac{d}{d t} [3^{\frac{1}{2}} t^{\frac{1}{2}}]$

Differentiate.

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Since $3^{\frac{1}{2}}$ is constant with respect to $t$ , the derivative of $3^{\frac{1}{2}} t^{\frac{1}{2}}$ with respect to $t$ is $3^{\frac{1}{2}} \frac{d}{d t} [t^{\frac{1}{2}}]$ .

$\sec^{2} (3^{\frac{1}{2}} t^{\frac{1}{2}}) (3^{\frac{1}{2}} \frac{d}{d t} [t^{\frac{1}{2}}])$

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

$\sec^{2} (3^{\frac{1}{2}} t^{\frac{1}{2}}) \cdot 3^{\frac{1}{2}} (\frac{1}{2} t^{\frac{1}{2} - 1})$

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

$\sec^{2} (3^{\frac{1}{2}} t^{\frac{1}{2}}) \cdot 3^{\frac{1}{2}} (\frac{1}{2} t^{\frac{1}{2} + \frac{- 1}{1} \cdot \frac{2}{2}})$

Write each expression with a common denominator of $2$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$\sec^{2} (3^{\frac{1}{2}} t^{\frac{1}{2}}) \cdot 3^{\frac{1}{2}} (\frac{1}{2} t^{\frac{1}{2} + \frac{- 1 \cdot 2}{1 \cdot 2}})$

Multiply $2$ by $1$ .

$\sec^{2} (3^{\frac{1}{2}} t^{\frac{1}{2}}) \cdot 3^{\frac{1}{2}} (\frac{1}{2} t^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}})$

Combine the numerators over the common denominator.

$\sec^{2} (3^{\frac{1}{2}} t^{\frac{1}{2}}) \cdot 3^{\frac{1}{2}} (\frac{1}{2} t^{\frac{1 - 1 \cdot 2}{2}})$

Simplify the numerator.

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Multiply $- 1$ by $2$ .

$\sec^{2} (3^{\frac{1}{2}} t^{\frac{1}{2}}) \cdot 3^{\frac{1}{2}} (\frac{1}{2} t^{\frac{1 - 2}{2}})$

Subtract $2$ from $1$ .

$\sec^{2} (3^{\frac{1}{2}} t^{\frac{1}{2}}) \cdot 3^{\frac{1}{2}} (\frac{1}{2} t^{\frac{- 1}{2}})$

Move the negative in front of the fraction.

$\sec^{2} (3^{\frac{1}{2}} t^{\frac{1}{2}}) \cdot 3^{\frac{1}{2}} (\frac{1}{2} t^{- \frac{1}{2}})$

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

$\sec^{2} (3^{\frac{1}{2}} t^{\frac{1}{2}}) \cdot 3^{\frac{1}{2}} \frac{t^{- \frac{1}{2}}}{2}$

Combine $\sec^{2} (3^{\frac{1}{2}} t^{\frac{1}{2}})$ and $\frac{t^{- \frac{1}{2}}}{2}$ .

$3^{\frac{1}{2}} \frac{\sec^{2} (3^{\frac{1}{2}} t^{\frac{1}{2}}) t^{- \frac{1}{2}}}{2}$

Combine $3^{\frac{1}{2}}$ and $\frac{\sec^{2} (3^{\frac{1}{2}} t^{\frac{1}{2}}) t^{- \frac{1}{2}}}{2}$ .

$\frac{3^{\frac{1}{2}} (\sec^{2} (3^{\frac{1}{2}} t^{\frac{1}{2}}) t^{- \frac{1}{2}})}{2}$

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

$\frac{3^{\frac{1}{2}} \sec^{2} (3^{\frac{1}{2}} t^{\frac{1}{2}})}{2 t^{\frac{1}{2}}}$

Find the Derivative – d/dt y=tan( square root of 3t)

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