设f(x)=cos1/x 则f'(2/π)=?

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设f(x)=cos1/x则f''(2/π)=?设f(x)=cos1/x则f''(2/π)=?设f(x)=cos1/x则f''(2/π)=?首先求f''(x)=?f(x)=cos(1/x)=-sin(1/x)*

设f(x)=cos1/x 则f'(2/π)=?
设f(x)=cos1/x 则f'(2/π)=?

设f(x)=cos1/x 则f'(2/π)=?
首先求f'(x)=?
f(x)=cos(1/x)=-sin(1/x)*(-1/x^2)=sin(1/x)*(1/x^2)
f'(2/π)=sin(π/2)*1/(π/2)^2=4/π^2