If `y=cosec(cotsqrtx),"then find "dy/dx`.

Answer

To find the derivative \( \frac{dy}{dx} \) of the function \( y = \csc(\cot(\sqrt{x})) \), we will use the chain rule and the derivatives of trigonometric functions. Let's go through the steps systematically.

### Step 1: Identify the function and its components
We have:
\[ y = \csc(\cot(\sqrt{x})) \]
To differentiate this, we need to apply the chain rule multiple times.

### Step 2: Differentiate the outer function
The derivative of \( \csc(u) \) with respect to \( u \) is:
\[ \frac{d}{du}(\csc(u)) = -\csc(u) \cot(u) \]
Let \( u = \cot(\sqrt{x}) \). Therefore, we have:
\[ \frac{dy}{du} = -\csc(\cot(\sqrt{x})) \cot(\cot(\sqrt{x})) \]

### Step 3: Differentiate the middle function
Next, we need to differentiate \( u = \cot(\sqrt{x}) \). The derivative of \( \cot(v) \) with respect to \( v \) is:
\[ \frac{d}{dv}(\cot(v)) = -\csc^2(v) \]
Let \( v = \sqrt{x} \). Thus,
\[ \frac{du}{dv} = -\csc^2(\sqrt{x}) \]

### Step 4: Differentiate the innermost function
Now we differentiate \( v = \sqrt{x} \):
\[ \frac{dv}{dx} = \frac{1}{2\sqrt{x}} \]

### Step 5: Apply the chain rule
Now we can combine these derivatives using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \left(-\csc(\cot(\sqrt{x})) \cot(\cot(\sqrt{x}))\right) \cdot \left(-\csc^2(\sqrt{x})\right) \cdot \left(\frac{1}{2\sqrt{x}}\right) \]

### Step 6: Simplify the expression
This simplifies to:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \csc(\cot(\sqrt{x})) \cot(\cot(\sqrt{x})) \csc^2(\sqrt{x}) \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \csc(\cot(\sqrt{x})) \cot(\cot(\sqrt{x})) \csc^2(\sqrt{x}) \]