Let `y =log_(e)sin(x^2),0ltxlepi/2`. The value of `dy/dx` at `x= sqrt(pi/2)` is

Options

A. 0
B. 1
C. `pi/4`
D. `sqrtpi`

Answer

To find the derivative \( \frac{dy}{dx} \) for the function \( y = \log_e(\sin(x^2)) \) and evaluate it at \( x = \sqrt{\frac{\pi}{2}} \), we will follow these steps:

### Step 1: Differentiate \( y \)

Given:
\[ y = \log_e(\sin(x^2)) \]

Using the chain rule, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{1}{\sin(x^2)} \cdot \frac{d}{dx}(\sin(x^2)) \]

### Step 2: Differentiate \( \sin(x^2) \)

Now we need to differentiate \( \sin(x^2) \) using the chain rule:
\[ \frac{d}{dx}(\sin(x^2)) = \cos(x^2) \cdot \frac{d}{dx}(x^2) = \cos(x^2) \cdot 2x \]

### Step 3: Substitute back into the derivative

Substituting this back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{\sin(x^2)} \cdot (2x \cos(x^2)) \]
Thus,
\[ \frac{dy}{dx} = \frac{2x \cos(x^2)}{\sin(x^2)} \]

### Step 4: Evaluate \( \frac{dy}{dx} \) at \( x = \sqrt{\frac{\pi}{2}} \)

Now we need to evaluate this derivative at \( x = \sqrt{\frac{\pi}{2}} \):
\[ \frac{dy}{dx} \bigg|_{x = \sqrt{\frac{\pi}{2}}} = \frac{2 \cdot \sqrt{\frac{\pi}{2}} \cdot \cos\left(\left(\sqrt{\frac{\pi}{2}}\right)^2\right)}{\sin\left(\left(\sqrt{\frac{\pi}{2}}\right)^2\right)} \]

Calculating \( \left(\sqrt{\frac{\pi}{2}}\right)^2 \):
\[ \left(\sqrt{\frac{\pi}{2}}\right)^2 = \frac{\pi}{2} \]

Thus, we have:
\[ \frac{dy}{dx} \bigg|_{x = \sqrt{\frac{\pi}{2}}} = \frac{2 \cdot \sqrt{\frac{\pi}{2}} \cdot \cos\left(\frac{\pi}{2}\right)}{\sin\left(\frac{\pi}{2}\right)} \]

### Step 5: Simplify using trigonometric values

We know that:
\[ \cos\left(\frac{\pi}{2}\right) = 0 \quad \text{and} \quad \sin\left(\frac{\pi}{2}\right) = 1 \]

Substituting these values:
\[ \frac{dy}{dx} \bigg|_{x = \sqrt{\frac{\pi}{2}}} = \frac{2 \cdot \sqrt{\frac{\pi}{2}} \cdot 0}{1} = 0 \]

### Final Answer

Thus, the value of \( \frac{dy}{dx} \) at \( x = \sqrt{\frac{\pi}{2}} \) is:
\[ \boxed{0} \]