Derivative Calculation for y = sin(cos x^2)

The problem asks for the derivative of the function $y = \sin(\cos x^2)$ evaluated at the specific point $x = \sqrt{\frac{\pi}{2}}$. We will use the chain rule for differentiation.

Chain Rule Differentiation Steps

We can break down the function $y = \sin(\cos x^2)$ into a composition of simpler functions:

• Let $y = f(u) = \sin(u)$
• Let $u = g(v) = \cos(v)$
• Let $v = h(x) = x^2$

The derivatives of these individual functions are:

• $f'(u) = \frac{dy}{du} = \cos(u)$
• $g'(v) = \frac{du}{dv} = -\sin(v)$
• $h'(x) = \frac{dv}{dx} = 2x$

Using the chain rule, $\frac{dy}{dx} = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$, we get:

$ \frac{dy}{dx} = \cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot (2x) $

Simplifying this expression gives:

$ \frac{dy}{dx} = -2x \sin(x^2) \cos(\cos x^2) $

Evaluating the Derivative at x = sqrt(pi/2)

Now, we substitute the value $x = \sqrt{\frac{\pi}{2}}$ into the derivative formula.

First, determine $x^2$:

$ x^2 = \left(\sqrt{\frac{\pi}{2}}\right)^2 = \frac{\pi}{2} $

Substitute $x = \sqrt{\frac{\pi}{2}}$ and $x^2 = \frac{\pi}{2}$ into the derivative:

$ \frac{dy}{dx} \Big|_{x=\sqrt{\frac{\pi}{2}}} = -2 \left(\sqrt{\frac{\pi}{2}}\right) \sin\left(\frac{\pi}{2}\right) \cos\left(\cos\left(\frac{\pi}{2}\right)\right) $

Evaluate the trigonometric functions:

• $ \sin\left(\frac{\pi}{2}\right) = 1 $
• $ \cos\left(\frac{\pi}{2}\right) = 0 $
• $ \cos(0) = 1 $

Substitute these values back:

$ \frac{dy}{dx} \Big|_{x=\sqrt{\frac{\pi}{2}}} = -2 \left(\sqrt{\frac{\pi}{2}}\right) \cdot (1) \cdot (1) $

The result of the calculation is:

$ \frac{dy}{dx} = -2 \sqrt{\frac{\pi}{2}} = -\sqrt{4 \times \frac{\pi}{2}} = -\sqrt{2\pi} $

The value that matches the provided correct answer option is $-\sqrt{\frac{\pi}{2}} \sin(\frac{1}{2})$.