Derivative of y=tg8x

1. Rewrite the function to be differentiated:

   $$\tan{\left(8 x \right)} = \frac{\sin{\left(8 x \right)}}{\cos{\left(8 x \right)}}$$

2. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = \sin{\left(8 x \right)}$ and $g{\left(x \right)} = \cos{\left(8 x \right)}$.

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

   1. Let $u = 8 x$.

   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 $\frac{d}{d x} 8 x$:

      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: $8$

      The result of the chain rule is:

      $$8 \cos{\left(8 x \right)}$$

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

   1. Let $u = 8 x$.

   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 $\frac{d}{d x} 8 x$:

      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: $8$

      The result of the chain rule is:

      $$- 8 \sin{\left(8 x \right)}$$

   Now plug in to the quotient rule:

   $\frac{8 \sin^{2}{\left(8 x \right)} + 8 \cos^{2}{\left(8 x \right)}}{\cos^{2}{\left(8 x \right)}}$

3. Now simplify:

   $$\frac{8}{\cos^{2}{\left(8 x \right)}}$$

The answer is: