If y = cos (sin {x^2}), then at x = sqrt {fra | vedclass

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(D) Given the function $y = \cos (\sin {x^2}).$ Applying the chain rule,we differentiate with respect to $x$: $\frac{dy}{dx} = -\sin (\sin {x^2}) \cdo

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$ 0 $

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(D) Given the function $y = \cos (\sin {x^2}).$ Applying the chain rule,we differentiate with respect to $x$: $\frac{dy}{dx} = -\sin (\sin {x^2}) \cdot \frac{d}{dx}(\sin {x^2}) $ $\frac{dy}{dx} = -\sin (\sin {x^2}) \cdot \cos {x^2} \cdot \frac{d}{dx}(x^2) $ $\frac{dy}{dx} = -\sin (\sin {x^2}) \cdot \cos {x^2} \cdot 2x $ Now,substitute $x = \sqrt {\frac{\pi }{2}} $ into the derivative: At $x = \sqrt {\frac{\pi }{2}} , x^2 = \frac{\pi }{2} $ Therefore,$\cos {x^2} = \cos \frac{\pi }{2} = 0 $ Substituting this value into the expression for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\sin (\sin \frac{\pi }{2}) \cdot 0 \cdot 2\sqrt {\frac{\pi }{2}} = 0 $ Thus,the correct option is $D$.

If $y = \cos (\sin {x^2}),$ then at $x = \sqrt {\frac{\pi }{2}} , \frac{dy}{dx} = $

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