Derivative of tg(t^3)

1. Rewrite the function to be differentiated:

   $$\tan{\left(t^{3} \right)} = \frac{\sin{\left(t^{3} \right)}}{\cos{\left(t^{3} \right)}}$$

2. Apply the quotient rule, which is:

   $$\frac{d}{d t} \frac{f{\left(t \right)}}{g{\left(t \right)}} = \frac{- f{\left(t \right)} \frac{d}{d t} g{\left(t \right)} + g{\left(t \right)} \frac{d}{d t} f{\left(t \right)}}{g^{2}{\left(t \right)}}$$

   $f{\left(t \right)} = \sin{\left(t^{3} \right)}$ and $g{\left(t \right)} = \cos{\left(t^{3} \right)}$.

   To find $\frac{d}{d t} f{\left(t \right)}$:

   1. Let $u = t^{3}$.

   2. The derivative of sine is cosine:

      $$\frac{d}{d u} \sin{\left(u \right)} = \cos{\left(u \right)}$$

   3. Then, apply the chain rule. Multiply by $\frac{d}{d t} t^{3}$:

      1. Apply the power rule: $t^{3}$ goes to $3 t^{2}$

      The result of the chain rule is:

      $$3 t^{2} \cos{\left(t^{3} \right)}$$

   To find $\frac{d}{d t} g{\left(t \right)}$:

   1. Let $u = t^{3}$.

   2. The derivative of cosine is negative sine:

      $$\frac{d}{d u} \cos{\left(u \right)} = - \sin{\left(u \right)}$$

   3. Then, apply the chain rule. Multiply by $\frac{d}{d t} t^{3}$:

      1. Apply the power rule: $t^{3}$ goes to $3 t^{2}$

      The result of the chain rule is:

      $$- 3 t^{2} \sin{\left(t^{3} \right)}$$

   Now plug in to the quotient rule:

   $\frac{3 t^{2} \sin^{2}{\left(t^{3} \right)} + 3 t^{2} \cos^{2}{\left(t^{3} \right)}}{\cos^{2}{\left(t^{3} \right)}}$

3. Now simplify:

   $$\frac{3 t^{2}}{\cos^{2}{\left(t^{3} \right)}}$$

The answer is:

$$\frac{3 t^{2}}{\cos^{2}{\left(t^{3} \right)}}$$