Derivative of y=ctg(x)+sqrt(x)

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

   1. There are multiple ways to do this derivative.

      Method #1

      1. Rewrite the function to be differentiated:

         $$\cot{\left(x \right)} = \frac{1}{\tan{\left(x \right)}}$$

      2. Let $u = \tan{\left(x \right)}$.

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

      4. Then, apply the chain rule. Multiply by $\frac{d}{d x} \tan{\left(x \right)}$:

         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 of the chain rule is:

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

      Method #2

      1. Rewrite the function to be differentiated:

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

      2. Apply the quotient rule, which is:

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

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

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

         1. The derivative of cosine is negative sine:

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

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

         1. The derivative of sine is cosine:

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

         Now plug in to the quotient rule:

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

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

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

2. Now simplify:

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

The answer is:

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