设f(x)对x可求导,求dy除以dx-搜答案-智慧作业帮

OK,我来说明,令g(x)=x^(1/2)由链式法则y=f(g(x))的导数为y'=f'(g(x))*g'(x)=f'(x^(1/2))*(x^(1/2))'=1/2*x^(-1/2)*f'(x^(1

记g(x)=f(x^2+sin^2x)+f(arctanx)=yg'(x)=f'(x^2+sin^2x)(2x+sin2x)+f'(arctanx)/(x2+1)dy/dx|x=0,即g'(0)代入得

根据复合函数求导法则dy/dx=[f(x^2)]'=f'(x^2)*(x^2)'=2xf'(x)

dy/dx=2xf'(x²))cosf(x²)再问：没有过程吗？再答：复合函数求导法则

用链式法则u=f(x)u'=f'(x)y=u³所以dy/dx==3u²*u'=3f²(x)*f'(x)再问：y=u³所以dy/dx==3u²*u'这个

曲线积分∫[f(x)+x]ydx+[f'(x)+sinx]dy与路径无关,那么：{[f(x)+x]y}‘y=[f'(x)+sinx]'xf''(x)+cosx=f(x)+xf''(x)-f(x)=x-

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)]

1）y'=f'(tanx)*(tanx)'=f'(tanx)*(secx)^22)y'=f'(x^2)*2x+f'(x)/f(x)

g'(x)=1/2/√{1+[sinf(x)]^2}*2sinf(x)cosf(x)f'(x)=sinf(x)cosf(x)f'(x)/√{1+[sinf(x)]^2}

可以把y看作f(e^x)与e^(f(x))相乘的函数,所以dy/dx=y'=[f(e^x)]'*e^(f(x))+f(e^x)*[e^(f(x))]'……………………(1)式其中[f(e^x)]'可看

做任务的.

(e^x)'=e^x,(x^e)'=e*x^(e-1),dy/dx=f'(e^x十x^e)*[e^x+e*x^(e-1)]

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)]/(

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

复合或者参数可以单变量显函数二阶导没有用除这一说再问：dx/dy不能当成f（x）/f(y)吗我汗，你大几的啊再答："dx/dy不能当成f（x）/f(y)吗我汗，你大几的啊"首先dx/dy或者dy/dx

dyf'(arcsin(1/x))—=－———————dxx√（x^2－1）

e^xy(y+xy')=2+6yy'y'=(ye^xy-2)/(6y-xe^xy)dy=dx(ye^xy-2)/(6y-xe^xy)

y=f[e^(-x)]y'=-f'[e^(-x)]*e^(-x)所以dy=-f'[e^(-x)]*e^(-x)dx