Question

57 - 64. second derivatives find y for the following functions.

1. y = x sin x
2. y = x² cos x
3. y = e^x sin x
4. y = 1/2 e^x cos x

Explanation:

1. For \(y = x\sin x\):

• Step 1: Find the first - derivative \(y'\) using the product rule \((uv)^\prime=u^\prime v + uv^\prime\), where \(u = x\) and \(v=\sin x\)
• \(u^\prime=\frac{d}{dx}(x)=1\), \(v^\prime=\frac{d}{dx}(\sin x)=\cos x\).
• \(y^\prime=\frac{d}{dx}(x\sin x)=1\times\sin x+x\times\cos x=\sin x + x\cos x\).
• Step 2: Find the second - derivative \(y''\) using the sum rule \((u + v)^\prime=u^\prime+v^\prime\) and product rule for \(x\cos x\)
• \(\frac{d}{dx}(\sin x)=\cos x\).
• For \(\frac{d}{dx}(x\cos x)\), using the product rule with \(u = x\), \(u^\prime = 1\), \(v=\cos x\), \(v^\prime=-\sin x\), we have \(\frac{d}{dx}(x\cos x)=1\times\cos x+x\times(-\sin x)=\cos x - x\sin x\).
• \(y''=\frac{d}{dx}(\sin x + x\cos x)=\cos x+(\cos x - x\sin x)=2\cos x - x\sin x\).

1. For \(y = x^{2}\cos x\):

• Step 1: Find the first - derivative \(y'\) using the product rule \((uv)^\prime=u^\prime v + uv^\prime\), where \(u = x^{2}\) and \(v=\cos x\)
• \(u^\prime=\frac{d}{dx}(x^{2}) = 2x\), \(v^\prime=\frac{d}{dx}(\cos x)=-\sin x\).
• \(y^\prime=\frac{d}{dx}(x^{2}\cos x)=2x\cos x - x^{2}\sin x\).
• Step 2: Find the second - derivative \(y''\) using the sum rule and product rule
• For \(\frac{d}{dx}(2x\cos x)\), using the product rule with \(u = 2x\), \(u^\prime = 2\), \(v=\cos x\), \(v^\prime=-\sin x\), we get \(\frac{d}{dx}(2x\cos x)=2\cos x+2x\times(-\sin x)=2\cos x - 2x\sin x\).
• For \(\frac{d}{dx}(x^{2}\sin x)\), using the product rule with \(u = x^{2}\), \(u^\prime = 2x\), \(v=\sin x\), \(v^\prime=\cos x\), we have \(\frac{d}{dx}(x^{2}\sin x)=2x\sin x+x^{2}\cos x\).
• \(y''=(2\cos x - 2x\sin x)-(2x\sin x+x^{2}\cos x)=2\cos x - 4x\sin x - x^{2}\cos x\).

1. For \(y = e^{x}\sin x\):

• Step 1: Find the first - derivative \(y'\) using the product rule \((uv)^\prime=u^\prime v + uv^\prime\), where \(u = e^{x}\) and \(v=\sin x\)
• \(u^\prime=\frac{d}{dx}(e^{x})=e^{x}\), \(v^\prime=\frac{d}{dx}(\sin x)=\cos x\).
• \(y^\prime=\frac{d}{dx}(e^{x}\sin x)=e^{x}\sin x+e^{x}\cos x=e^{x}(\sin x+\cos x)\).
• Step 2: Find the second - derivative \(y''\) using the product rule for \(e^{x}(\sin x+\cos x)\)
• Let \(u = e^{x}\), \(u^\prime=e^{x}\), \(v=\sin x+\cos x\), \(v^\prime=\cos x-\sin x\).
• \(y''=e^{x}(\sin x+\cos x)+e^{x}(\cos x - \sin x)=2e^{x}\cos x\).

1. For \(y=\frac{1}{2}e^{x}\cos x\):

• Step 1: Find the first - derivative \(y'\) using the product rule \((uv)^\prime=u^\prime v + uv^\prime\), where \(u=\frac{1}{2}e^{x}\) and \(v = \cos x\)
• \(u^\prime=\frac{1}{2}e^{x}\), \(v^\prime=-\sin x\).
• \(y^\prime=\frac{1}{2}e^{x}\cos x-\frac{1}{2}e^{x}\sin x=\frac{1}{2}e^{x}(\cos x - \sin x)\).
• Step 2: Find the second - derivative \(y''\) using the product rule for \(\frac{1}{2}e^{x}(\cos x - \sin x)\)
• Let \(u=\frac{1}{2}e^{x}\), \(u^\prime=\frac{1}{2}e^{x}\), \(v=\cos x - \sin x\), \(v^\prime=-\sin x-\cos x\).
• \(y''=\frac{1}{2}e^{x}(\cos x - \sin x)+\frac{1}{2}e^{x}(-\sin x - \cos x)=-e^{x}\sin x\).

Answer:

• For \(y = x\sin x\), \(y''=2\cos x - x\sin x\).
• For \(y = x^{2}\cos x\), \(y''=2\cos x - 4x\sin x - x^{2}\cos x\).
• For \(y = e^{x}\sin x\), \(y''=2e^{x}\cos x\).
• For \(y=\frac{1}{2}e^{x}\cos x\), \(y''=-e^{x}\sin x\).