y=cos√x-2求dy-搜答案-神马作文网

y=(1+x^4)^cos x,求dy

(1+X4)COSx先求导数再乘以dx就行了

设y=cos^2(x+1),求dy

楼上的少写了“-”和“dx”吧dy=2cos(x+1)•[-sin(x+1)]dx=-sin2(x+1)dx

y= (cos x)^x,求dy/dx

最好的办法是求对数：lny=xlncosx,两边求导数得：y'/y=lncosx-x(sinx/cosx)=lncosx-xtanx所以：y'=y(lncosx-xtanx)=(cosx)^x(lnc

再答：再问：dy再答：��ѽ��再问：΢��dy再问：û�ĵ�再答：

e^(-y^2)=-cos(x^2)第一次求导后：e^(-y^2)*（-2y）*（dy/dx）=2x*sin(x^2)dy/dx=-[x*sin(x^2)]/[y*e^(-y^2)]

解y=cos[ln(x²+e^(-x)]y'={cos[ln(x²+e^(-x)]}'=-sin[ln(x²+e^(-x)][ln(x²+e^(-x)]'(x&

两边对x求导：dy/dxsinx+ycosx-sin(x-y)(1-dy/dx)=0,将x=π/2带入已知方程得到y,再把x、y带入上式求得结果再问：x=π/2带入已知方程得到y。。。我算不出这个y

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).

应用复合函数求导方法,y′sinx+ycosx+（1+y′）sin（x+y）=0,（sinx+sin（x+y））y′+ycosx+sin（x+y）=0,y′=-（ycosx+sin（x+y））/（si

cos(x+y)+y=1 求dy/dx

我算的结果和你的一样,也是y'=sin(x+y)/1-sin(x+y)应该是书上写错了.在说xsin(x+y)中的x从何而来?找不到它的来源啊.不管是对cos(),还是对y求导都不会出现xsin()这

正确答案：【-sin根号x/（2根号x）+2的x次方ln2】dx

dy=[-sin(√x)*1/2*x^(-1/2)-e^(-2x)*(-2)]dx=[1/2sin(√x)x^(-1/2)+2e^(-2x)]dx

y'=4*cos^3(2x+π/3)*[cos(2x+π/3)]'*(2x+π/3)'=-8*cos^3(2x+π/3)*sin(2x+π/3)

dy/dx=-2cosxsinx-5x的4次方所以dy=(-sin2x-5x的4次方)dx

dy=cos√xdx-e^(-2x)dx=sin√x*√xdx-e^(-2x)(-2x)dx=[√x(sin√x)/2x+2e^(-2x)]dx很高兴为您解答,【数学之美】团队为您答题.请点击下面的【

y=cos√x+2＾xy'=-sin√x/(2√x)+ln2*2＾xdy=[-sin√x/(2√x)+ln2*2＾x]dx

∵x=cos(y/x)==>1=-sin(y/x)*(xy'-y)/x^2(等式两端对x求导)==>xy'-y=-x^2/sin(y/x)==>xy'=y-x^2/sin(y/x)==>y'=y/x-

两边取导数得：[ysin3x-cos(y-x)]`=0即：y`sin3x+3ycos3x-sin(y-x)(y`-1)=0即：y`sin3x-y`sin(y-x)=sin(y-x)-3xcos3x所以

y=x^2x,求dy

y=x^(2x)lny=2xlnx(1/y)dy=(2+2lnx)dxdy=x^(2x).(2+2lnx)dx

两边求导：y'sinx+ycosx+sin(x+y)*(1+y')=0令x=0,y=π/2：π/2+1+y'=0y'=-(π/2+1)dy=-(π/2+1)dx