设fu为可导函数y=f(x³)则dy dx
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lim(x->0)[f(1)-f(1-x)]/2x=1lim(x->0)[f(1)-f(1-x)]/x=2即曲线在(1,f(1))处切线斜率为2
这是一个复合函数y=f(u(x))的求导,按下面公式:y'=f'(u)*u'(x)所以导数为:f'(x^2)*2x
1.dy/dx=f'(x^3)*3x^22.dy/dx=f'(e^x+x^e)*(e^x+ex^(e-1))3.dy/dx=f'(e^x)*(e^x)e^f(x)+f(e^x)[e^f(x)]*f'(
lim[f(1)-f(1-2x)]/2x=-1(中间是减号吧,否则有错)所以f'(1)=-1即y=f(x)在点(1,f(1))处的斜率为-1.再问:是减号谢谢咯~
由题,设1-x=t,则lim[1+f(t)]/2(1-t)=-1,t趋向于1因此可知,limf(t)=-1,t趋向于1;又因为f(x)可导,故其连续,故f(1)=-1.同时,上极限式可变为:lim[f
lim[f(1)-f(1-x)]/2x=-2化为:lim[f(1-x)-f(1)]/(-x)=-4因此有f'(1)=-4
lim[f(1)-f(1-2x)]/2x=lim[f(1)-f(1-2x)]/(0-2x)=f'(1)=-1∴曲线y=f(x)在点(1,f(1))处的斜率是-1再问:f'(1)=-1怎么来的?再答:f
dy/dx=cos{f[sinf(x)]}*{f[sinf(x)]}'=cos{f[sinf(x)]}*f‘[sinf(x)]*[sinf(x)]’=cos{f[sinf(x)]}*f‘[sinf(x
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)]
令u=x^yv=y^xdz/dx=dz/du*du/dx+dz/dv*dv/dx=df/du*y*x^(y-1)+df/dv*lny*y^xdz/dy=dz/du*du/dy+dz/dv*dv/dy=
1)y'=f'(tanx)*(tanx)'=f'(tanx)*(secx)^22)y'=f'(x^2)*2x+f'(x)/f(x)
函数f(x)可导,设其导函数为g(x)dy/dx=df(x^2)/dx=g(x^2)*dx^2/dx=2x*g(x^2)
dz=f'x(x/y)dx+f'y(x/y)dy=[f'(x/y)/y]dx+f'(x/y)(-x/y²)dy
两边求微分:d(x^y+y^x)=d(f(x^2+y^2))对x^y可以这么看:先把X看成常数,对Y求微分相当于a^Y,再把Y看成常数对X求微分相当于X^a.那么就好用公式了如下:d(x^y)=X^Y
两边对x求导得:2yy'*f(x)+y^2f'(x)+f(x)+xf'(x)=2x得:y'=[2x-xf'(x)-y^2f'(x)]/(2yf(x)]dy=[2x-xf'(x)-y^2f'(x)]/(
∂z/∂x=-((∂f/∂x)*y*2x)/f^2∂z/∂y=1/f+2y2*(∂f/∂y)/f^21/
这个是复合函数的求导问题dy/dx=f'(sin^2x)*(sin^2x)'+f'(cos^2x)*(cos^2x)'=f'(sin^2x)(2sinx*cosx)+f'(cos^2x)*(-2cos
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)