Derivative of y=tgx*sin(2x-5)

1. Apply the product rule:

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

   $f{\left(x \right)} = \tan{\left(x \right)}$; to find $\frac{d}{d x} f{\left(x \right)}$:

   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:

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

   $g{\left(x \right)} = \sin{\left(2 x - 5 \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Let $u = 2 x - 5$.

   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 $\frac{d}{d x} \left(2 x - 5\right)$:

      1. Differentiate $2 x - 5$ term by term:

         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: $2$

         2. The derivative of the constant $-5$ is zero.

         The result is: $2$

      The result of the chain rule is:

      $$2 \cos{\left(2 x - 5 \right)}$$

   The result is: $\frac{\left(\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}\right) \sin{\left(2 x - 5 \right)}}{\cos^{2}{\left(x \right)}} + 2 \cos{\left(2 x - 5 \right)} \tan{\left(x \right)}$

2. Now simplify:

   $$\frac{\sin{\left(2 x - 5 \right)} + \frac{\sin{\left(4 x - 5 \right)}}{2} + \frac{\sin{\left(5 \right)}}{2}}{\cos^{2}{\left(x \right)}}$$

The answer is:

$$\frac{\sin{\left(2 x - 5 \right)} + \frac{\sin{\left(4 x - 5 \right)}}{2} + \frac{\sin{\left(5 \right)}}{2}}{\cos^{2}{\left(x \right)}}$$