Find the derivative of `cos( sin x^(2)) at x = sqrt(pi/2)`

Explanation

To find the derivative of the function \( y = \cos(\sin(x^2)) \) at \( x = \sqrt{\frac{\pi}{2}} \), we will follow these steps:

### Step 1: Differentiate the function
We start with the function:
\[ y = \cos(\sin(x^2)) \]
To find the derivative \( y' \), we will use the chain rule. The chain rule states that if you have a composite function \( f(g(x)) \), then the derivative is given by:
\[ f'(g(x)) \cdot g'(x) \]

### Step 2: Differentiate the outer function
The outer function is \( f(u) = \cos(u) \), where \( u = \sin(x^2) \). The derivative of \( \cos(u) \) is:
\[ f'(u) = -\sin(u) \]
So,
\[ \frac{dy}{du} = -\sin(\sin(x^2)) \]

### Step 3: Differentiate the inner function
Now we differentiate the inner function \( g(x) = \sin(x^2) \). The derivative of \( \sin(v) \) where \( v = x^2 \) is:
\[ g'(x) = \cos(x^2) \cdot \frac{d}{dx}(x^2) = \cos(x^2) \cdot 2x \]

### Step 4: Combine the derivatives
Using the chain rule, we combine the derivatives:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\sin(\sin(x^2)) \cdot (2x \cos(x^2)) \]
Thus,
\[ \frac{dy}{dx} = -2x \cos(x^2) \sin(\sin(x^2)) \]

### Step 5: Evaluate the derivative at \( x = \sqrt{\frac{\pi}{2}} \)
Now we substitute \( x = \sqrt{\frac{\pi}{2}} \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{x = \sqrt{\frac{\pi}{2}}} = -2\left(\sqrt{\frac{\pi}{2}}\right) \cos\left(\left(\sqrt{\frac{\pi}{2}}\right)^2\right) \sin\left(\sin\left(\left(\sqrt{\frac{\pi}{2}}\right)^2\right)\right) \]
Calculating \( \left(\sqrt{\frac{\pi}{2}}\right)^2 \):
\[ \left(\sqrt{\frac{\pi}{2}}\right)^2 = \frac{\pi}{2} \]
Now substituting this into the cosine function:
\[ \cos\left(\frac{\pi}{2}\right) = 0 \]
Thus, the entire expression becomes:
\[ \frac{dy}{dx} \bigg|_{x = \sqrt{\frac{\pi}{2}}} = -2\left(\sqrt{\frac{\pi}{2}}\right) \cdot 0 \cdot \sin\left(\sin\left(\frac{\pi}{2}\right)\right) = 0 \]

### Final Answer
The derivative of \( \cos(\sin(x^2)) \) at \( x = \sqrt{\frac{\pi}{2}} \) is:
\[ \boxed{0} \]