91Ó°ÊÓ

To differentiate the given function, y = x * cos(x^2), we will apply the product rule. The product rule states that if you have a function of a product of two functions, e.g., u(x) * v(x), the derivative of the product is: $$(u * v)' = u' * v + u * v'$$ In our case, u = x, and v = cos(x^2). First, find the derivative of u and v: $$u' = \frac{d}{dx} (x) = 1$$ $$v' = \frac{d}{dx} (\cos x^2) = -\sin x^2 * \frac{d}{dx} (x^2) = -2x\sin x^2$$ Now, apply the product rule: $$\frac{dy}{dx} = u' * v + u * v' = 1 * \cos x^2 + x * (-2x\sin x^2)$$ $$\frac{dy}{dx} = \cos x^2 - 2x^2\sin x^2$$

We differentiate the first derivative \(\frac{dy}{dx}\) to get the second derivative \(\frac{d^2y}{dx^2}\). $$\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\cos x^2 - 2x^2\sin x^2\right)$$ Now we need to differentiate the functions inside the parentheses separately. For the first term, we apply the chain rule since it's a composite function. $$\frac{d}{dx}(\cos x^2) = -\sin x^2 * \frac{d}{dx}(x^2) = -2x\sin x^2$$ For the second term, we'll need to apply both the chain rule and the product rule: $$\frac{d}{dx}(-2x^2\sin x^2) = -(2x^2)' * \sin x^2 - 2x^2 * (\sin x^2)'$$ $$= -4x\sin x^2 - 2x^2 * (\cos x^2 * (x^2)')$$ $$= -4x\sin x^2 - 4x^3\cos x^2$$ Combine both derivatives to get the final answer: $$\frac{d^2y}{dx^2} = -2x\sin x^2 - 4x\sin x^2 - 4x^3\cos x^2$$ $$\frac{d^2y}{dx^2} = -6x\sin x^2 - 4x^3\cos x^2$$ The second derivative of the given function is \(\frac{d^2y}{dx^2} = -6x\sin x^2 - 4x^3\cos x^2\).