f'(sin x)=1-cos x 求f''(x)
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f''(sinx)=1-cosx求f''''(x)f''(sinx)=1-cosx求f''''(x)f''(sinx)=1-cosx求f''''(x)f''(sinx)=1-根号(1-sin^2x)f''(x)=1-根号(
f'(sin x)=1-cos x 求f''(x)
f'(sin x)=1-cos x 求f''(x)
f'(sin x)=1-cos x 求f''(x)
f'(sinx)=1-根号(1-sin^2x)
f'(x)=1-根号(1-x^2)
f"(x)=-(-2x)/(2根号(1-x^2))
=x/根号(1-x^2)
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