91Ó°ÊÓ

To find the first derivative, \(y'\), we'll use the product rule since we have two functions, \(x\) and \(\sin x\), being multiplied together. The product rule states that if we have a function \(y = uv\), where \(u\) and \(v\) are functions of \(x\), then the derivative of \(y\) is given by: $$y' = u'v + uv'$$ Let \(u = x\) and \(v = \sin x\). Then, we have \(u' = 1\) and \(v' = \cos x\). Now, we apply the product rule to find the first derivative: $$y' = (1)(\sin x) + (x)(\cos x) = \sin x + x\cos x$$

Now, we need to find the second derivative \(y''\). We can notice that \(y'\) is still a product of two functions, namely \(y' = x\cos x + 1\cdot \sin x\). We can apply the product rule again to find \(y''\). For the first term \(x\cos x\), let \(u_1 = x\) and \(v_1 = \cos x\), so we have \(u_1' = 1\) and \(v_1' = -\sin x\). Applying the product rule, we get: $$(x\cos x)' = (1)(\cos x) + (x)(-\sin x) = \cos x - x\sin x$$ For the second term \(\sin x\), we have a single function and can find its derivative directly: $$(\sin x)' = \cos x$$ Now, we can find \(y''\) by summing the derivatives of both terms: $$y'' = (\cos x - x\sin x) + \cos x = 2\cos x - x\sin x$$ So, the second derivative of the given function is: $$y'' = 2\cos x - x\sin x$$