设f(x)=sin(2x+3),求f(x)的n次导,并求出f(-3/2)的9次导

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设f(x)=sin(2x+3),求f(x)的n次导,并求出f(-3/2)的9次导设f(x)=sin(2x+3),求f(x)的n次导,并求出f(-3/2)的9次导设f(x)=sin(2x+3),求f(x

设f(x)=sin(2x+3),求f(x)的n次导,并求出f(-3/2)的9次导
设f(x)=sin(2x+3),求f(x)的n次导,并求出f(-3/2)的9次导

设f(x)=sin(2x+3),求f(x)的n次导,并求出f(-3/2)的9次导

n为奇数时
f(x)^n=sin(2x+3) * (2)^n * (-1)^(n+1)
f(-3/2)的9次导为0