设f'(cosx)=cos2x,求f'(sinx)

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设f'(cosx)=cos2x,求f'(sinx)设f'(cosx)=cos2x,求f'(sinx)设f'(cosx)=cos2x,求f'(sinx)f''(cosx)=cos(2x)=2cos^2x-

设f'(cosx)=cos2x,求f'(sinx)
设f'(cosx)=cos2x,求f'(sinx)

设f'(cosx)=cos2x,求f'(sinx)
f'(cosx)=cos(2x)=2cos^2x-1
f'(x)=2x^2-1
f'(sinx)=2sin^2x-1=-cos(2x)

  • cos2x=2(cosx)^2-1,令x=cosx,则f'(x)=2x^2-1

  • f'(sinx)=2(sinx)^2-1=-cos2x