ysin2x1求dy

苗瑶詹1015y=ln(2√x - 1) 求dy -
禹秀初13224362571 ______ y=ln(2√x-1) dy=dln(2√x-1)=1/(2√x-1) d(2√x-1)=1/(2√x-1) d(2√x)=2/(2√x-1) *1/(2√x) dx=1/(2x-√x)dx

苗瑶詹1015Y=Incos2x 求dy -
禹秀初13224362571 ______ ∵ (lnx)'=1/x (cosx)'=-sinx ∴ dy=d(Incos2x) =1/(cos2x) *(cos2x)' =1/(cos2x) *(-sin2x)*(2x)' =-sin2x/cos2x *2*dx =-2tan2x dx

苗瑶詹1015设函数求dy 急要 感激不尽设函数y=y^2sinx,则dy= -
禹秀初13224362571 ______[答案] ∵y=y²sinx ∴dy=sinxd(y²)+y²d(sinx) ==>dy=2ysinxdy+y²cosxdx ==>(1-2ysinx)dy=y²cosxdx 故 dy=y²cosxdx/(1-2ysinx).

苗瑶詹1015求解微分方程的通项:dy - ysin^2xdx=0 -
禹秀初13224362571 ______[答案] dy-ysin^2xdx=0 dy=ysin^2xdx dy/y=sin^2xdx dy/y=(1/2-1/2cos2x)dx d(lny)=d(x/2-1/4sin2x) lny=x/2-1/4sin2x+C1 则y=C*e^(x/2-1/4sin2x)

苗瑶詹1015y=sinx^2+2x,求dy,第一个2是平方的意思 -
禹秀初13224362571 ______ 很简单:dy=(sin2x+2)dx

苗瑶詹1015设y=cos(2x^2+x+1) 求dy/dx. 过程 -
禹秀初13224362571 ______ y=cos(2x^2+x+1) →dy/dx=-sin(2x^2+x+1)·(2x^2+x+1)' ∴dy/dx=-(4x+1)sin(2x^2+x+1).

苗瑶詹1015y=√(x^2+1),求dy -
禹秀初13224362571 ______ dy=d√(x^2+1), =d(x^2+1)^(1/2) =(1/2)*(x^2+1)^(-1/2)*d(x^2+1) =[x/√(x^2+1)]dx

苗瑶詹1015y=sin(e∧x+1),求dy -
禹秀初13224362571 ______ y=sin(e∧x+1) dy=cos(e^x+1)d(e^x+1) =e^xcos(e^x+1)dx

苗瑶詹1015设y=cos^2(x+1),求dy -
禹秀初13224362571 ______[答案] 楼上的少写了“-”和“dx”吧 dy=2cos(x+1)•[-sin(x+1)]dx=-sin2(x+1)dx

苗瑶詹1015已知y=xsinx 1/2cosx 求dy/dx|x=兀/4 -
禹秀初13224362571 ______ y=xsinx+1/2cosx=xsinx+(1/2)*secx dy/dx=sinx+xcosx+(1/2)*secxtanx dy/dx|(x=π/4)=sin(π/4)+(π/4)*cos(π/4)+(1/2)*sec(π/4)*tan(π/4)=√2/2+√2π/8+(1/2)*√2*1=(1+π/8)*√2