y=f(ln3x)求dy
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f(x)=2ln(3x)+8x=2lnx+8x+2ln3.lim[f(1-2x)-f(1)]/x=2lim[ln(1-2x)-8x]/x(0/0)=2lim[-2/(1-2x)-8]/1=-20.再问
dy/dx=2xf'(x²))cosf(x²)再问:没有过程吗?再答:复合函数求导法则
应该等于2xf'(x^2),看成复合函数就行了……
y=f[(x-1)/(x+1)],f'(x)=arctanx^2,求dy/dx,dy两边对x求导:dy/dx=f'[(x-1)/(x+1)]*2/(x+1)^2=arctan[(x-1)/(x+1)]
y=f(x^3)+f(sinx)复合函数求导:y'=f'(x^3)(x^3)'+f'(sinx)(sinx)'=3x^2f'(x^3)+cosxf'(sinx)所以dy/dx=3x^2f'(x^3)+
你们都不全对:求导结果你是对的但是x取值不对因为它已经给出表达式y=f[(x-1)/(x+1)],所以x≠-1所以x应为(x≠±1)
dy/dx=(x+y)/(x-y),dy/dx=(1+y/x)/(1-y/x)u=y/x,dy/dx=u+xdu/dx(1+u)/(1-u)=u+xdu/dxdx/x=(1-u)du/(1+u^2)l
复合函数的求导法则:如果u=g(x)在点x可导,而y=f(u)在点u=g(x)可导,则复合函数y=f[g(x)]在点x可导,且其导数为dy/dx=f'(u)g'(x)或dy/dx=(dy/du)(du
令u=x+arctanx,则u'=1+1/(1+x^2)则y=f^2(u)dy/dx=2f(u)f'(u)u'=2f(u)f'(u)[1+1/(x+x^2)]
dy/dx=(1/y)+k=(1+ky)/y得到ydy/(1+ky)=dxydy/(1+ky)=(1/k)*(1+ky-1)dy/(1+ky)=dx得到(1+ky-1)dy/(1+ky)=kdx=(1
三个变量,两个方程,所以任何一个变量都能表示其余两个变量,偏微分可以写成微分 对f求x的偏微分,=>其中fi分别是f对第i个未知数的偏导数对g求x的偏微分,=>
再问:��Ҫ��cosxô再答:��Ȼ�Ǹ��Ϻ�����˳��������������
两边对x求导:f'(x²+y²)(2x+2yy')=y'解得:y'=2xf'(x²+y²)/[1-2yf'(x²+y²)]希望可以帮到你,如
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
y=ln(ln2(ln3x))y'=[1/(ln2(ln3x))]*(ln2(ln3x))'=[1/(ln2(ln3x))]*[2ln(ln3x)]*(1/ln3x)*(3ln2x)*(1/x)
1.y=ln[ln²(ln³x)]y'=[ln²(ln³x)]'/ln²(ln³x)=2ln(ln³x)*[ln(ln³
恩,dy=df(sinx)=f'(sinx)*d(sinx)=f'(sinx)*cosxdx结果到这里应该可以了吧?
dyf'(arcsin(1/x))—=-———————dxx√(x^2-1)
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