y=(sinx+cosx)^(1/2)
y'=1/2(sinx+cosx)^(-1/2)・(sinx+cosx)'
=(cosx-sinx)/2(sinx+cosx)^(1/2)
です。
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y=(sinx+cosx)^(1/2)
y'=1/2(sinx+cosx)^(-1/2)・(sinx+cosx)'
=(cosx-sinx)/2(sinx+cosx)^(1/2)
です。
Users viewing this question
are also looking at these questions 😉
