57 - 64. second derivatives find y for the following functions.\n57. y = x sin x\n58. y = x² cos x\n59. y =…
57 - 64. second derivatives find y for the following functions.\n57. y = x sin x\n58. y = x² cos x\n59. y = e^x sin x\n60. y = 1/2 e^x cos x
Answer
- 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).
- 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)
- 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).
- 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)
- 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).
- 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)
- 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).
- 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)
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).