Derivative of y=2cosx-3tgx

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

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

      1. The derivative of cosine is negative sine:

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

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

   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:

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

      2. Apply the quotient rule, which is:

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

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

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

         1. The derivative of sine is cosine:

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

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

         1. The derivative of cosine is negative sine:

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

         Now plug in to the quotient rule:

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

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

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

2. Now simplify:

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

The answer is:

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