Find derivative of `f(x) = sqrt cosx` using first principle.

Answer

To find the derivative of the function \( f(x) = \sqrt{\cos x} \) using the first principle of derivatives, we will follow these steps:

### Step 1: Set up the definition of the derivative
The derivative of a function \( f(x) \) using the first principle is defined as:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Substitute the function into the definition
Substituting \( f(x) = \sqrt{\cos x} \) into the formula gives:
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{\cos(x+h)} - \sqrt{\cos x}}{h} \]

### Step 3: Identify the indeterminate form
As \( h \) approaches 0, both \( \sqrt{\cos(x+h)} \) and \( \sqrt{\cos x} \) approach the same value, leading to a \( \frac{0}{0} \) indeterminate form. Thus, we need to simplify this expression.

### Step 4: Rationalize the numerator
To simplify the expression, we multiply the numerator and denominator by the conjugate of the numerator:
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{\cos(x+h)} - \sqrt{\cos x}}{h} \cdot \frac{\sqrt{\cos(x+h)} + \sqrt{\cos x}}{\sqrt{\cos(x+h)} + \sqrt{\cos x}} \]
This gives:
\[ f'(x) = \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h \left( \sqrt{\cos(x+h)} + \sqrt{\cos x} \right)} \]

### Step 5: Use the cosine difference identity
Using the identity for the difference of cosines:
\[ \cos(x+h) - \cos x = -2 \sin\left(\frac{x+h+x}{2}\right) \sin\left(\frac{h}{2}\right) \]
We can substitute this into our limit:
\[ f'(x) = \lim_{h \to 0} \frac{-2 \sin\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h \left( \sqrt{\cos(x+h)} + \sqrt{\cos x} \right)} \]

### Step 6: Simplify the limit
As \( h \) approaches 0, \( \sin\left(\frac{h}{2}\right) \) approaches \( \frac{h}{2} \). Thus, we can rewrite the limit:
\[ f'(x) = \lim_{h \to 0} \frac{-2 \sin\left(x + \frac{h}{2}\right) \cdot \frac{h}{2}}{h \left( \sqrt{\cos(x+h)} + \sqrt{\cos x} \right)} \]
This simplifies to:
\[ f'(x) = \lim_{h \to 0} \frac{-\sin\left(x + \frac{h}{2}\right)}{\sqrt{\cos(x+h)} + \sqrt{\cos x}} \]

### Step 7: Evaluate the limit
As \( h \) approaches 0, \( \sin\left(x + \frac{h}{2}\right) \) approaches \( \sin x \) and \( \sqrt{\cos(x+h)} + \sqrt{\cos x} \) approaches \( 2\sqrt{\cos x} \):
\[ f'(x) = \frac{-\sin x}{2\sqrt{\cos x}} \]

### Final Answer
Thus, the derivative of \( f(x) = \sqrt{\cos x} \) is:
\[ f'(x) = -\frac{\sin x}{2\sqrt{\cos x}} \]