设函数y=2 lnx,则dy=-搜答案-拍题作业网

方程两边同时求x对y的导：y+xdy/dx+1/x+2ydy/dx=0,dy/dx=-(y+1/x)/(x+2y),dy=-(y+1/x)dx/(x+2y)

设y=x+lnx,则dx/dy=

dx/dy=x'(y)y=x+lnx,对等号两边求导∴1=x'+x'/x∴x'(1+1/x)=1∴dx/dy=x'=x/(x+1)

是y=linx²么dy/dx=1/x².（x²）'=1/x².(2x)=2x/x²dy/dx|x=1=2*1/1²=2

y'=e^x(tanx+lnx)+e^x((secx)^2+1/x)=e^x(tanx+lnx+(secx)^2+1/x)dy=[e^x(tanx+lnx+(secx)^2+1/x)]dx

已知：x^(y²)+y²lnx=4即：e^[lnx^(y²)]+y²lnx=4即：e^[y²×lnx]+y²lnx=4两边求导得到：e^[(

symsx;y=atan((x^2-1)^(1/2))-log(x)/((x^2-1)^(1/2))y=atan((x^2-1)^(1/2))-log(x)/(x^2-1)^(1/2)>>diff(y

两天同乘以e^(∫P(x)dx)则左边变成[ye^(∫P(x)dx)]',右边是Q(x)e^(∫P(x)dx)所以ye^(∫P(x)dx)=∫Q(x)e^(∫P(x)dx)dx+Cy=e^(-∫P(x

x=y*e^(-y)故dx/dy=e^(-y)+y*(-e^(-y))=(1-y)*e^(-y)故dy/dx=e^y/(1-y)再问：是吧dy/dx看成分数的是吧？

y=lnx^2,dy/dx=

(2ylnx*y

商法则(f/g)'=(f'g-g'f)/g^2dy/dx=[(2+lnx)'*x-(2+lnx)*x']/x^2=[1/x*x-(2+lnx)*1]/x^2=[-1-lnx]/x^2代入x=elne=

(1+1/x)dx其实也就是求导dy=f'(x)(f(x)的导数）dx

y'=[f(lnx)]'e^f(x)+f(lnx)[e^f(x)]'=f'(lnx)(lnx)'e^f(x)+f(lnx)e^f(x)[f(x)]'=f'(lnx)e^f(x)/x+f(lnx)e^f

再问：你是用对数求导法么？第一步右边那个怎么写得出来的再答：

【(lnx-1)/(lnx²)】'=[1/lnx-1/(lnx)²]'=[(lnx)^(-1)-(lnx)^(-2)]'=(-1/x)(lnx)^(-2)+(1/x)2(lnx)^

y'=[(lnx)'sinx-lnx*(sinx)']/(sinx)^2=(sinx*1/x-lnx*cosx)/(sinx)^2所以dy=(sinx*1/x-lnx*cosx)/(sinx)^2dx

原式y=sinx^2+2xdy/dx=2x·cosx^2+2

设函数y=x^x,则dy=

y=x^x那么lny=xlnx两边对x求导得：y'/y=lnx+1所以y'=y(1+lnx)即dy/dx=y(1+lnx)=x^x*(1+lnx)所以dy=x^x*(1+lnx)dx如果不懂,祝学习愉

DY=[1/(x根号下lnx)-2e^(-2x)]Dx

y=(e^2x)/lnx,求dy

DY=[1/(x根号下lnx)-2e^(-2x)]Dx