Use delta method to find the derivatives of the following : `sqrt(sinx)`

Answer

To find the derivative of \( y = \sqrt{\sin x} \) using the delta method, we will follow these steps:

### Step 1: Define the function
Let
\[ y = f(x) = \sqrt{\sin x} \]

### Step 2: Apply the delta method
According to the delta method, the derivative \( \frac{dy}{dx} \) can be expressed as:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Substituting \( f(x) \):
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{\sqrt{\sin(x+h)} - \sqrt{\sin x}}{h} \]

### Step 3: Rationalize the numerator
To simplify the expression, we can multiply and divide by the conjugate:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{\left(\sqrt{\sin(x+h)} - \sqrt{\sin x}\right)\left(\sqrt{\sin(x+h)} + \sqrt{\sin x}\right)}{h\left(\sqrt{\sin(x+h)} + \sqrt{\sin x}\right)} \]
This gives us:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h\left(\sqrt{\sin(x+h)} + \sqrt{\sin x}\right)} \]

### Step 4: Use the sine difference identity
Using the identity \( \sin a - \sin b = 2 \cos\left(\frac{a+b}{2}\right) \sin\left(\frac{a-b}{2}\right) \):
\[ \sin(x+h) - \sin x = 2 \cos\left(\frac{(x+h)+x}{2}\right) \sin\left(\frac{(x+h)-x}{2}\right) = 2 \cos\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right) \]
Substituting this back into our limit:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{2 \cos\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h\left(\sqrt{\sin(x+h)} + \sqrt{\sin x}\right)} \]

### Step 5: Simplify the limit
We know that \( \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \to 1 \) as \( h \to 0 \), hence:
\[ \sin\left(\frac{h}{2}\right) \approx \frac{h}{2} \text{ as } h \to 0 \]
Thus, we can rewrite:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{2 \cos\left(x + \frac{h}{2}\right) \cdot \frac{h}{2}}{h\left(\sqrt{\sin(x+h)} + \sqrt{\sin x}\right)} = \lim_{h \to 0} \frac{\cos\left(x + \frac{h}{2}\right)}{\sqrt{\sin(x+h)} + \sqrt{\sin x}} \]

### Step 6: Evaluate the limit
As \( h \to 0 \), \( \cos\left(x + \frac{h}{2}\right) \to \cos x \) and \( \sqrt{\sin(x+h)} \to \sqrt{\sin x} \):
\[ \frac{dy}{dx} = \frac{\cos x}{\sqrt{\sin x} + \sqrt{\sin x}} = \frac{\cos x}{2\sqrt{\sin x}} \]

### Final Result
Thus, the derivative of \( y = \sqrt{\sin x} \) is:
\[ \frac{dy}{dx} = \frac{\cos x}{2\sqrt{\sin x}} \]