sec x / (1+x^2),求dy/dx忘打了--- y= (sec x) / (1+x^2)

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sec x / (1+x^2),求dy/dx忘打了--- y= (sec x) / (1+x^2)

sec x / (1+x^2),求dy/dx忘打了--- y= (sec x) / (1+x^2)
sec x / (1+x^2),求dy/dx
忘打了--- y= (sec x) / (1+x^2)

sec x / (1+x^2),求dy/dx忘打了--- y= (sec x) / (1+x^2)
利用(u/v)'=[u'v-v'u]/v²
dy/dx=y'=[(1+x²)secxtanx-2xsecx]/(1+x²)²={secx[tanx(1+x²)-2x]}/(1+x²)²