If `y = log(tan sqrt(x))`, then the value of `(dy)/(dx)` is

Options

A. `1/(2sqrtx)`
B. `("sec"^(2)sqrt(x))/(sqrt(x) tan x)`
C. `2 "sec"^(2) sqrt(x)`
D. `("sec"^(2)sqrt(x))/(2sqrt(x)tan sqrt(x))`

Answer

To find the derivative \( \frac{dy}{dx} \) for the function \( y = \log(\tan(\sqrt{x})) \), we will use the chain rule and the properties of logarithmic and trigonometric functions.

### Step-by-Step Solution:

1. **Identify the function**:
\[ y = \log(\tan(\sqrt{x})) \]

2. **Differentiate using the chain rule**:
The derivative of \( \log(u) \) with respect to \( x \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = \tan(\sqrt{x}) \).
\[ \frac{dy}{dx} = \frac{1}{\tan(\sqrt{x})} \cdot \frac{d}{dx}(\tan(\sqrt{x})) \]

3. **Differentiate \( \tan(\sqrt{x}) \)**:
We apply the chain rule again. The derivative of \( \tan(v) \) is \( \sec^2(v) \cdot \frac{dv}{dx} \), where \( v = \sqrt{x} \).
\[ \frac{d}{dx}(\tan(\sqrt{x})) = \sec^2(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x}) \]

4. **Differentiate \( \sqrt{x} \)**:
The derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \).
\[ \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} \]

5. **Combine the derivatives**:
Now substitute back into the derivative of \( \tan(\sqrt{x}) \):
\[ \frac{d}{dx}(\tan(\sqrt{x})) = \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \]

6. **Substitute back into \( \frac{dy}{dx} \)**:
Now we can substitute this back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{\tan(\sqrt{x})} \cdot \left(\sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}\right) \]

7. **Simplify the expression**:
We can rewrite \( \sec^2(\sqrt{x}) \) as \( \frac{1}{\cos^2(\sqrt{x})} \) and \( \tan(\sqrt{x}) \) as \( \frac{\sin(\sqrt{x})}{\cos(\sqrt{x})} \):
\[ \frac{dy}{dx} = \frac{1}{\tan(\sqrt{x})} \cdot \frac{1}{2\sqrt{x}} \cdot \sec^2(\sqrt{x}) = \frac{\sec^2(\sqrt{x})}{2\sqrt{x} \tan(\sqrt{x})} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{\sec^2(\sqrt{x})}{2\sqrt{x} \tan(\sqrt{x})} \]