Derivative of tan³(4x)

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

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

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

   1. Rewrite the function to be differentiated:

      $$\tan{\left(4 x \right)} = \frac{\sin{\left(4 x \right)}}{\cos{\left(4 x \right)}}$$

   2. Apply the quotient rule, which is:

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

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

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

      1. Let $u = 4 x$.

      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 $\frac{d}{d x} 4 x$:

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

            1. Apply the power rule: $x$ goes to $1$

            So, the result is: $4$

         The result of the chain rule is:

         $$4 \cos{\left(4 x \right)}$$

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

      1. Let $u = 4 x$.

      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 $\frac{d}{d x} 4 x$:

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

            1. Apply the power rule: $x$ goes to $1$

            So, the result is: $4$

         The result of the chain rule is:

         $$- 4 \sin{\left(4 x \right)}$$

      Now plug in to the quotient rule:

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

   The result of the chain rule is:

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

4. Now simplify:

   $$\frac{12 \tan^{2}{\left(4 x \right)}}{\cos^{2}{\left(4 x \right)}}$$

The answer is:

$$\frac{12 \tan^{2}{\left(4 x \right)}}{\cos^{2}{\left(4 x \right)}}$$