题目

求导y=根号下(2x+3)(x-1)/(2x-1)(x+2)

提问时间：2021-01-13

答案

y=√{(2x+3)(x-1)/[(2x-1)(x+2)]}
=√[(2x+3)(x-1)]/√[(2x-1)(x+2)]=u/v
y'=(u/v)'=[u'v-uv']/v^2
={1/2*1/√[(2x+3)(x-1)]*[2*(x-1)+(2x+3)*1]*√[(2x-1)(x+2)]-√[(2x+3)(x-1)]*1/2*1/√[(2x-1)(x+2)]*[2*(x+2)+(2x-1)*1]}/{√[(2x-1)(x+2)]}^2
=(4x+1)]*√[(2x-1)(x+2)]/√[(2x+3)(x-1)]-(4x+3)]*√[(2x+3)(x-1)]/√[(2x-1)(x+2)]
=[(4x+1)(2x-1)(x+2)-(4x+2)(2x+3)(x-1)]/{2(2x-1)(x+2)*√[(2x+3)(x-1)(2x-1)(x+2)]}
=(4x^2+6x+7)/{2(2x-1)(x+2)*√[(2x+3)(x-1)(2x-1)(x+2)]}