Find the second order derivative of the funct | vedclass

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(A) Let $y = x \cdot \cos x$. First,we find the first derivative using the product rule: $\frac{dy}{dx} = \frac{d}{dx}(x) \cdot \cos x + x \cdot \frac

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$-(x \cos x + 2 \sin x)$

Vedclass · Answer

(A) Let $y = x \cdot \cos x$. First,we find the first derivative using the product rule: $\frac{dy}{dx} = \frac{d}{dx}(x) \cdot \cos x + x \cdot \frac{d}{dx}(\cos x)$ $= 1 \cdot \cos x + x \cdot (-\sin x) = \cos x - x \sin x$. Now,we find the second derivative: $\frac{d^2y}{dx^2} = \frac{d}{dx}(\cos x - x \sin x) = \frac{d}{dx}(\cos x) - \frac{d}{dx}(x \sin x)$. Using the product rule for the second term: $\frac{d}{dx}(x \sin x) = \frac{d}{dx}(x) \cdot \sin x + x \cdot \frac{d}{dx}(\sin x) = 1 \cdot \sin x + x \cos x = \sin x + x \cos x$. Substituting this back: $\frac{d^2y}{dx^2} = -\sin x - (\sin x + x \cos x) = -\sin x - \sin x - x \cos x = -(x \cos x + 2 \sin x)$.

Find the second order derivative of the function $f(x) = x \cdot \cos x$.

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