Derivative of y=tg^5(3x^4-13)

1. Let $u = \tan{\left(3 x^{4} - 13 \right)}$.

2. Apply the power rule: $u^{5}$ goes to $5 u^{4}$

3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \tan{\left(3 x^{4} - 13 \right)}$:

   1. Rewrite the function to be differentiated:

      $$\tan{\left(3 x^{4} - 13 \right)} = \frac{\sin{\left(3 x^{4} - 13 \right)}}{\cos{\left(3 x^{4} - 13 \right)}}$$

   2. Apply the quotient rule, which is:

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

      $f{\left(x \right)} = \sin{\left(3 x^{4} - 13 \right)}$ and $g{\left(x \right)} = \cos{\left(3 x^{4} - 13 \right)}$.

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

      1. Let $u = 3 x^{4} - 13$.

      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 $\frac{d}{d x} \left(3 x^{4} - 13\right)$:

         1. Differentiate $3 x^{4} - 13$ term by term:

            1. The derivative of a constant times a function is the constant times the derivative of the function.

               1. Apply the power rule: $x^{4}$ goes to $4 x^{3}$

               So, the result is: $12 x^{3}$

            2. The derivative of the constant $-13$ is zero.

            The result is: $12 x^{3}$

         The result of the chain rule is:

         $$12 x^{3} \cos{\left(3 x^{4} - 13 \right)}$$

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

      1. Let $u = 3 x^{4} - 13$.

      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 $\frac{d}{d x} \left(3 x^{4} - 13\right)$:

         1. Differentiate $3 x^{4} - 13$ term by term:

            1. The derivative of a constant times a function is the constant times the derivative of the function.

               1. Apply the power rule: $x^{4}$ goes to $4 x^{3}$

               So, the result is: $12 x^{3}$

            2. The derivative of the constant $-13$ is zero.

            The result is: $12 x^{3}$

         The result of the chain rule is:

         $$- 12 x^{3} \sin{\left(3 x^{4} - 13 \right)}$$

      Now plug in to the quotient rule:

      $\frac{12 x^{3} \sin^{2}{\left(3 x^{4} - 13 \right)} + 12 x^{3} \cos^{2}{\left(3 x^{4} - 13 \right)}}{\cos^{2}{\left(3 x^{4} - 13 \right)}}$

   The result of the chain rule is:

   $$\frac{5 \left(12 x^{3} \sin^{2}{\left(3 x^{4} - 13 \right)} + 12 x^{3} \cos^{2}{\left(3 x^{4} - 13 \right)}\right) \tan^{4}{\left(3 x^{4} - 13 \right)}}{\cos^{2}{\left(3 x^{4} - 13 \right)}}$$

4. Now simplify:

   $$\frac{60 x^{3} \tan^{4}{\left(3 x^{4} - 13 \right)}}{\cos^{2}{\left(3 x^{4} - 13 \right)}}$$

The answer is: