If f(x)`=cos (sin x^(2))`, then `f'(x)` at `x=sqrt(pi/2)` is

Options

A. `-1`
B. 1
C. 0
D. 2

Answer

To find \( f'(x) \) where \( f(x) = \cos(\sin(x^2)) \) and evaluate it at \( x = \sqrt{\frac{\pi}{2}} \), we will follow these steps:

### Step 1: Differentiate \( f(x) \)
Using the chain rule, we differentiate \( f(x) = \cos(\sin(x^2)) \).

\[ f'(x) = -\sin(\sin(x^2)) \cdot \frac{d}{dx}(\sin(x^2)) \]

### Step 2: Differentiate \( \sin(x^2) \)
Next, we need to differentiate \( \sin(x^2) \) using the chain rule again.

\[ \frac{d}{dx}(\sin(x^2)) = \cos(x^2) \cdot \frac{d}{dx}(x^2) = \cos(x^2) \cdot 2x \]

### Step 3: Substitute back into \( f'(x) \)
Now we substitute this back into our expression for \( f'(x) \):

\[ f'(x) = -\sin(\sin(x^2)) \cdot \cos(x^2) \cdot 2x \]

### Step 4: Evaluate \( f'(\sqrt{\frac{\pi}{2}}) \)
Now we need to evaluate \( f' \) at \( x = \sqrt{\frac{\pi}{2}} \):

1. Calculate \( x^2 \):
\[ x^2 = \left(\sqrt{\frac{\pi}{2}}\right)^2 = \frac{\pi}{2} \]

2. Calculate \( \sin(x^2) \):
\[ \sin\left(\frac{\pi}{2}\right) = 1 \]

3. Calculate \( \cos(x^2) \):
\[ \cos\left(\frac{\pi}{2}\right) = 0 \]

4. Substitute these values into \( f'(\sqrt{\frac{\pi}{2}}) \):
\[ f'\left(\sqrt{\frac{\pi}{2}}\right) = -\sin(1) \cdot 0 \cdot 2\left(\sqrt{\frac{\pi}{2}}\right) \]

Since one of the factors is 0, we have:

\[ f'\left(\sqrt{\frac{\pi}{2}}\right) = 0 \]

### Final Answer
Thus, \( f'(\sqrt{\frac{\pi}{2}}) = 0 \).

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