If ` y= sec ( tan sqrt x ) ,then (dy)/(dx) =`

Options

A. `(tan sqrt xsec ^(2) sqrtx sec (tan sqrtx ))/( 2sqrtx) `
B. ` (sec^(2)sqrt ( x)sec (tan sqrt x ) tan (tan sqrt x))/( 2sqrtx ) `
C. ` (sec ^(2) sqrt x sec ( tan sqrt x) tan (tan sqrt x))/( sqrt x) `
D. ` (tan sqrtx sec ^(2) sqrtx sec (tan sqrtx ))/( sqrtx) `

Answer

To find the derivative of the function \( y = \sec(\tan(\sqrt{x})) \), we will apply the chain rule of differentiation step by step.

### Step 1: Identify the outer and inner functions
Let:
- \( u = \tan(\sqrt{x}) \)
- \( y = \sec(u) \)

### Step 2: Differentiate \( y \) with respect to \( u \)
Using the derivative of the secant function:
\[ \frac{dy}{du} = \sec(u) \tan(u) \]

### Step 3: Differentiate \( u \) with respect to \( x \)
Now we need to differentiate \( u = \tan(\sqrt{x}) \). We apply the chain rule again:
- Let \( v = \sqrt{x} \)
- Then \( u = \tan(v) \)

First, we differentiate \( u \) with respect to \( v \):
\[ \frac{du}{dv} = \sec^2(v) \]

Next, we differentiate \( v \) with respect to \( x \):
\[ \frac{dv}{dx} = \frac{1}{2\sqrt{x}} \]

Now, we can find \( \frac{du}{dx} \):
\[ \frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} = \sec^2(v) \cdot \frac{1}{2\sqrt{x}} = \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \]

### Step 4: Combine the derivatives using the chain rule
Now we can combine the derivatives using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \sec(u) \tan(u) \cdot \frac{du}{dx} = \sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x})) \cdot \left( \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \right) \]

### Final Answer
Thus, the derivative is:
\[ \frac{dy}{dx} = \sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x})) \cdot \frac{\sec^2(\sqrt{x})}{2\sqrt{x}} \]