设函数y=y(x)由方程-热点-学帮网

设函数y=y(x)由方程(x+y)^(1/x)=y所确定,则dy/dx=?设函数y=y(x)由方程(x+y)^(1/x)=y所确定,则dy/dx=?(答案没有ln,可是我老带着ln)ln(x+y)=x·lny(1+y‘)/(x+y)=lny

设函数y=y(x)由方程tany=y-x确定,则dy=?tany=y-x两边同时对x求导得：sec²y*y'=y'-1所以y'=-1/(sec²y-1)=-1/tan²y所以dy/dx=-1/tan²

设函数y由方程lny+x/y=0确定,求dy/dxlny+x/y=0等式两边求导：y'*1/y+1/y+x*y'(-1/y²)=0(1/y-x/y²)y'=-1/y∴y'=(-1/y)/(1/y-x/y²)=-

设函数y=y(x)由方程y=tan(x+y)所确定求y''y'=（1+y'）/cos^2（x+y）解出y'=-1/sin^2（x+y）在求导得y''=2（1+y'）*cos（x+y）/sin^3（x+y）然后把y'=-1/sin^2（x+y

设由下列方程确定隐函数y=f(x),求y''.方程是y=1+(xe)^y

设函数y=y(x)由方程y+e^(x+y)=2x确定,求dx/dy分别对y求导,求左边为1+【e^(x+y）×（dx/dy+1）】右边为2×dx/dy推的dx/dy：自己算下,没得草稿纸.y+e^(x+y)=2xy'+(1+y')e^(x+

设函数y=y(x)由方程cos(x+y)+y=1确定,求dy/dx由隐函数微分法可得：－sin(x＋y)(1＋y′)＋y′＝0－sin(x＋y)＋[1－sin(x＋y)]y′＝0∴y′＝sin(x＋y)／[1－sin(x＋y)]．-sin(

设隐函数y=y(x)由方程x^y-e^y=sin(xy)所确定,求dy化为：e^(ylnx)-e^y=sin(xy)两边对x求导：e^(ylnx)(y'lnx+y/x)-y'e^y=cos(xy)(y+xy')y'[lnxe^(ylnx)-

设函数y=y(x)由方程e^y+xy+e^x=0确定,求y''(0)e^y+xy+e^x=0两边同时对x求导得：e^y·y'+y+xy'+e^x=0得y'=-(y+e^x)/(x+e^y)y''=-[(y'+e^x)(x+e^y)-(y+e

设函数y=y(x)由方程y+x=e的xy次方确定,求y'(0)y+x=e^xyx=0y=1y'+1=(y+xy')e^xy1-ye^xy=y'(xe^xy-1)y'=(1-ye^xy)/(xe^xy-1)y'(0)=(1-1)/(-1)=0

设y(x)由方程e^y-e^x=xy所确定的隐函数求y'y'(0)e^y-e^x=xy两边求导,得e^y*y'-e^x=y+xy'(e^y-x)y'=(e^x+y)所以y'=(e^x+y)/(e^y-x)x=0时,e^y-e^0=0,则e^

设函数y=y(x)由方程lny=tan(xy)所确定,求dy左右对x求导有y'/y=sec²(xy)(y+xy')整理有y'=y²/(cos(xy)-xy)所以dy=(y²/(cos(xy)-xy))dxdy=

设函数Y=F(X)由方程Y=1+YX决定,求DY/DXY=1+YXy=1/(1-x)DY/DX=(-1)(1-x)^(-2)*(-1)=1/(1-x)^2【【不清楚,再问；满意,祝你好运开☆!】】

设函数x=x(y)由方程y^3+x+y=4确定,求dx设函数x=x(y)由方程y^3+x+y=4确定,求dx|x=1方程y^3+x+y=4变形x=-y^3-y+4dx=(-3y^2-1)dy方程两边同取微分，可以得到：dx=-(1+3y^2

设函数y=y(x)由方程xy+e^y=1所确定,求y"(0)xy+e^y=1e^y(0)=1y(0)=0xy'+y+e^yy'=00+y(0)+y'(0)=0y'(0)=0xy''+y'+y'+e^yy''+(y')^2e^y=00+2y'

设函数y=y（x）由方程e^y+xy=e所确定,求y’（0）两边对x求导数,得y'*e^y+y+xy'=0,在原方程中令x=0可得y=1,因此,将x=0,y=1代入上式可得y'+1=0,即y'(0)=-1.

设函数y=y（x）由方程e*y+xy=e所决定,求y'（0）.对方程求x的导数:(e*y)'+(xy)'=(e)'ey'+(y+xy')=0y'=-y/(e+x)当x=0,y=1,y'=-y/(e+x)=-1/(e+0)=-1/e显然当x=

设函数y=y(x)由方程sin(x²y)+ln(2x-y)=0所确定,则曲线y=y(x)在点(0.-1)处的切线方程为（0,-1）在曲线上,是切点对x求导cos(x²y)*(2xy+x²*y')+1/(2x-y

设函数y=y(x)由方程y=f(x^2+y^2)+f(x+y)确定,且y(0)=2,f(x)是可导函数,f'(2)=1/2,f'(4)=1,则f'(0)的值y=f(x²+y²)+f(x+y)y'=f'(x²+y