Derivative of 2sinx/cos^3x

1. Apply the quotient rule, which is:

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

   $f{\left(x \right)} = 2 \sin{\left(x \right)}$ and $g{\left(x \right)} = \cos^{3}{\left(x \right)}$.

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

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

      1. The derivative of sine is cosine:

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

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

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

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

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

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

      1. The derivative of cosine is negative sine:

         $$\frac{d}{d x} \cos{\left(x \right)} = - \sin{\left(x \right)}$$

      The result of the chain rule is:

      $$- 3 \sin{\left(x \right)} \cos^{2}{\left(x \right)}$$

   Now plug in to the quotient rule:

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

2. Now simplify:

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

The answer is:

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