Derivative of y(x)=-3cos2x+3tg4x

1. Differentiate $- 3 \cos{\left(2 x \right)} + 3 \tan{\left(4 x \right)}$ term by term:

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

      1. Let $u = 2 x$.

      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 $\frac{d}{d x} 2 x$:

         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: $2$

         The result of the chain rule is:

         $$- 2 \sin{\left(2 x \right)}$$

      So, the result is: $6 \sin{\left(2 x \right)}$

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

      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:

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

      So, the result is: $\frac{3 \left(4 \sin^{2}{\left(4 x \right)} + 4 \cos^{2}{\left(4 x \right)}\right)}{\cos^{2}{\left(4 x \right)}}$

   The result is: $\frac{3 \left(4 \sin^{2}{\left(4 x \right)} + 4 \cos^{2}{\left(4 x \right)}\right)}{\cos^{2}{\left(4 x \right)}} + 6 \sin{\left(2 x \right)}$

2. Now simplify:

   $$6 \sin{\left(2 x \right)} + \frac{12}{\cos^{2}{\left(4 x \right)}}$$

The answer is:

$$6 \sin{\left(2 x \right)} + \frac{12}{\cos^{2}{\left(4 x \right)}}$$