设f '(sinx^2)=cos2x+tanx^2 (0
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设f '(sinx^2)=cos2x+tanx^2 (0
![设f '(sinx^2)=cos2x+tanx^2 (0](/uploads/image/z/18605111-23-1.jpg?t=%E8%AE%BEf+%27%28sinx%5E2%29%3Dcos2x%2Btanx%5E2+%280)
f '(sinx^2)=cos2x+tanx^2=1-2sinx^2+(sinx^2)/(1-sinx^2)
therefore
f'(x)=1-2x+x/(1-x)=1/(1-x)-2x
f(x)=-ln(1-x)-x^2
therefore
f'(x)=1-2x+x/(1-x)=1/(1-x)-2x
f(x)=-ln(1-x)-x^2
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