求微分 y=cos根号x-2的x次方,求dy-搜答案-智慧作业帮

我来试试吧...z=e^xy*cos(x+y)Z'x=ye^xycos(x+y)-e^xysin(x+y)Z'y=xe^xycos(x+y)-e^xysin(x+y)故dZ=[ye^xycos(x+y

隐函数求导设z=x²y²-cos(xy)dy/dx=-(δz/δx)/(δz/δy)=-(2xy²+ysin(xy))/(2x²y+xsin(xy))=-y/x

-sin(xy)[ydx+xdy]=2xy^2*dx+x^2*2ydy-sin(xy)ydx-sin(xy)xdy=2xy^2*dx+2x^2*ydy-2x^2*ydy-sin(xy)xdy=2xy^

再答：再问：dy再答：��ѽ��再问：΢��dy再问：û�ĵ�再答：

-((2x)/(1-x^2))dx;(-E^-x-Sin[3+x])dx;2Cos[2x]dx

微分的乘积法则和链式法则学过吗?这两个都是微分基本的法则,做这道题时都会用到.y=e^(-x)cos(3-x)dy/dx={d[e^(-x)]/dx}cos(3-x)+e^(-x){d[cos(3-x

求x=y^y的微分

如果对x求导,则ln|x|=yln|y|,1/x=y'/y+yy'/y=y'/y+y',.对数求导法.如果对y求导,则ln|x|=yln|y|,x'/x=ln|y|+y/y,x'=y^y(1+ln|y

微分dy=[-e的（-x）cos(3-x)+e的（-x）sin（3-x）]dx要想求导的话就直接把dx移到前边去就好了.

symsx>>y=log(x+sqrt(1+x^2));>>simple(diff(y)ans=1/(1+x^2)^(1/2)>>y=log(2*x+sqrt(1+x^2));>>simple(dif

y＝ln[x+√(1+x²)]∴y'＝[x+√(1+x²)]'/[x+√(1+x²)]＝[1+x/√(1+x²)]/[x+√(1+x²)]＝[x+√(

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(1)y=(2x^3-3x^2+3)(√x+1/x)=2x^(7/2)-3x^(3/2)+3x^(1/2)+2x^2-3x+3/x,　dy=[7x^(5/2)-(9/2)x^(1/2)+(3/2)x^

1.d(cosx)^2=2cosx(-sinx)dx=-sin2xdx2.dsin(x²-1)=cos(x²-1)d(x²-1)=cos(x²-1)×2xdx=

y=arcsin√(1-x^2)y'=-x/(|x|√(1-x^2))∴dy=-xdx/(|x|√(1-x^2))当x>0dy=-dx/√(1-x^2)当x

y=[ln(1-x)^2]^2y'=2[ln(1-x)^2]*[ln(1-x)^2]'=2[ln(1-x)^2]*[2ln(1-x)]'=2[ln(1-x)^2]*2*1/(1-x)=4*[ln(1-

z=1/2*ln(x^2+y^2+4)Z'x=1/2*1/(x^2+y^2+4)*(2x)=x/(x^2+y^2+4)Z'y=1/2*1/(x^2+y^2+4)*(2y)=y/(x^2+y^2+4)所

y=x^2(cosx+√x),dy=[2x(cosx+√x)+x²(-sinx+1/2*1/√x)]dx=[2xcosx-x²sinx+2x√x+1/2*x√x]dx=[x(2co

y'=2e^2xcos(e^2x)把y看成复合函数sint,t=e^m,m=2x.复合函数求导,等于三个分别求导的积

y=1/x+√x=x^(-1)+x^(1/2)∴y'=(-1)*x^(-2)+(1/2)*x^(-1/2)=-1/x^2+1/(2√x)