Mathematics
Senior High

この問題の解き方分かる方お願いします。

微分

Answers

y=sin⁻¹(2x-1)

u=2x-1と置換すると、

y'=(sin⁻¹u)'*(2x-1)'

={1/√(1-u²)}*2

uを元に戻して、

y'=〔2/√{1-(2x-1)²}〕

={2/√(-2x²+4x)}

=〔2/√{-2x(x-2)}〕

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