Derivative of y=x^3tgx

1. Apply the product rule:

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

   $f{\left(x \right)} = x^{3}$; to find $\frac{d}{d x} f{\left(x \right)}$:

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

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

   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:

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

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

2. Now simplify:

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

The answer is:

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