Derivative of exp(4*x)/((2*tan(4*x)))

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = e^{4 x}$ and $g{\left(x \right)} = 2 \tan{\left(4 x \right)}$.

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

   1. Let $u = 4 x$.

   2. The derivative of $e^{u}$ is itself.

   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:

      $$4 e^{4 x}$$

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

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

   Now plug in to the quotient rule:

   $\frac{- \frac{2 \left(4 \sin^{2}{\left(4 x \right)} + 4 \cos^{2}{\left(4 x \right)}\right) e^{4 x}}{\cos^{2}{\left(4 x \right)}} + 8 e^{4 x} \tan{\left(4 x \right)}}{4 \tan^{2}{\left(4 x \right)}}$

2. Now simplify:

   $$\left(\frac{2}{\tan{\left(4 x \right)}} - \frac{2}{\sin^{2}{\left(4 x \right)}}\right) e^{4 x}$$

The answer is: