Derivative of (tg2x)*sqrt(3x^2-5x+1)

1. Apply the product rule:

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

   $f{\left(x \right)} = \tan{\left(2 x \right)}$; to find $\frac{d}{d x} f{\left(x \right)}$:

   1. Rewrite the function to be differentiated:

      $$\tan{\left(2 x \right)} = \frac{\sin{\left(2 x \right)}}{\cos{\left(2 x \right)}}$$

   2. Apply the quotient rule, which is:

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

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

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

      1. Let $u = 2 x$.

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

         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:

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

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

      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:

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

      Now plug in to the quotient rule:

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

   $g{\left(x \right)} = \sqrt{3 x^{2} - 5 x + 1}$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Let $u = 3 x^{2} - 5 x + 1$.

   2. Apply the power rule: $\sqrt{u}$ goes to $\frac{1}{2 \sqrt{u}}$

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(3 x^{2} - 5 x + 1\right)$:

      1. Differentiate $3 x^{2} - 5 x + 1$ 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: $x^{2}$ goes to $2 x$

            So, the result is: $6 x$

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

            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: $5$

            So, the result is: $-5$

         3. The derivative of the constant $1$ is zero.

         The result is: $6 x - 5$

      The result of the chain rule is:

      $$\frac{6 x - 5}{2 \sqrt{3 x^{2} - 5 x + 1}}$$

   The result is: $\frac{\left(6 x - 5\right) \tan{\left(2 x \right)}}{2 \sqrt{3 x^{2} - 5 x + 1}} + \frac{\left(2 \sin^{2}{\left(2 x \right)} + 2 \cos^{2}{\left(2 x \right)}\right) \sqrt{3 x^{2} - 5 x + 1}}{\cos^{2}{\left(2 x \right)}}$

2. Now simplify:

   $$\frac{6 x^{2} - 10 x + \frac{\left(6 x - 5\right) \sin{\left(4 x \right)}}{4} + 2}{\sqrt{3 x^{2} - 5 x + 1} \cos^{2}{\left(2 x \right)}}$$

The answer is: