Derivative of (x+secx)(x-tanx)

1. Apply the product rule:

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

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

   1. Differentiate $x + \sec{\left(x \right)}$ term by term:

      1. Apply the power rule: $x$ goes to $1$

      2. Rewrite the function to be differentiated:

         $$\sec{\left(x \right)} = \frac{1}{\cos{\left(x \right)}}$$

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

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

      5. 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:

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

      The result is: $\frac{\sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} + 1$

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

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

      1. Apply the power rule: $x$ goes to $1$

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

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

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

2. Now simplify:

   $$\frac{- \left(x + \frac{1}{\cos{\left(x \right)}}\right) \sin^{2}{\left(x \right)} + \left(x - \tan{\left(x \right)}\right) \left(\sin{\left(x \right)} + \cos^{2}{\left(x \right)}\right)}{\cos^{2}{\left(x \right)}}$$

The answer is:

$$\frac{- \left(x + \frac{1}{\cos{\left(x \right)}}\right) \sin^{2}{\left(x \right)} + \left(x - \tan{\left(x \right)}\right) \left(\sin{\left(x \right)} + \cos^{2}{\left(x \right)}\right)}{\cos^{2}{\left(x \right)}}$$