91Ó°ÊÓ

To find the first derivative \(\frac{dy}{dx}\), we use the product rule for the function \(y = x \cos x^{2}\). The product rule states that if \(y = u(x)v(x)\), then \(\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)\). In our case, \(u(x) = x\) and \(v(x) = \cos x^{2}\), so: $$\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) = \frac{d}{dx}\left(x\right)\left(\cos x^{2}\right) + x\frac{d}{dx}\left(\cos x^2\right)$$

Now, we need to calculate the derivatives: \(\frac{d}{dx}\left(x\right)\) and \(\frac{d}{dx}\left(\cos x^2\right)\). The derivative of \(x\) is \(1\). For the derivative of \(\cos x^2\), we use the chain rule, which states if \(y = f(g(x))\), then \(\frac{dy}{dx} = f'(g(x))g'(x)\). Here, \(f(x) = \cos x\) and \(g(x) = x^2\): $$\frac{d}{dx}\left(\cos x^2\right) = -\sin x^2\left(2x\right) = -2x\sin x^2$$ Now, we plug these derivatives back into the expression for \(\frac{dy}{dx}\): $$\frac{dy}{dx} = (1)(\cos x^2) + x(-2x\sin x^2) = \cos x^2 - 2x^2\sin x^2$$

To find the second derivative \(\frac{d^2y}{dx^2}\), we take the derivative of \(\frac{dy}{dx}\) with respect to \(x\). So, we need to calculate the derivatives: \(\frac{d}{dx}\left(\cos x^2\right)\) and \(\frac{d}{dx}\left(-2x^2\sin x^2\right)\). We have already calculated the derivative for \(\cos x^2\) in Step 2: $$\frac{d}{dx}\left(\cos x^2\right) = -2x\sin x^2$$ For the derivative of \(-2x^2\sin x^2\), we again use the product rule. Let \(u(x) = -2x^2\) and \(v(x) = \sin x^2\). Applying the product rule: $$\frac{d}{dx}\left(-2x^2\sin x^2\right) = \left(\frac{d}{dx}(-2x^2)\right)\left(\sin x^2\right) + \left(-2x^2\right)\left(-2x\sin x^2\right)$$ Now, calculate the derivatives: \(\frac{d}{dx}\left(-2x^2\right)\) and \(\frac{d}{dx}\left(\sin x^2\right)\). The derivative of \(-2x^2\) is \(-4x\). For the derivative of \(\sin x^2\), we use the chain rule again, with \(f(x) = \sin x\) and \(g(x) = x^2\): $$\frac{d}{dx}\left(\sin x^2\right) = \cos x^2\left(2x\right) = 2x\cos x^2$$ Plugging the derivatives back into the expression for the second derivative: $$\frac{d^2y}{dx^2} = \left(-4x\right)\left(\sin x^2\right) -2x^2\left(-2x\sin x^2\right) = -4x\sin x^2 + 4x^3\sin x^2$$

Finally, we simplify the expression for the second derivative: $$\frac{d^2y}{dx^2} = -4x\sin x^2\left(1 - x^2\right)$$ So, the second derivative of the given function \(y = x \cos x^{2}\) is: $$\frac{d^2y}{dx^2} = -4x\sin x^2\left(1 - x^2\right)$$