不是2(x)^(1/2)f'(x^1/2)
追答不是
是对X求导 不是对根号下X求导
设f(x)是可导函数,y=f(根号下x),则dy\/dx=?,是y和x分别求导,然后二者...
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\/2))
f(x),g(x)可导,求y=f[g(2^x)]函数的导数dy\/dx
如图所示:
设f(x),g(x)可导,y=f(e^x)e^g(x),求dy\/dx
这个题目包含2个知识点:(1)乘积的微分,y=f(x)g(x),则dy\/dx=f'(x)g(x)+f(x)g'(x)(2)复合函数的微分,链式法则,y=f(g(x)),dy\/dx=f'(g(x))g'(x)根据以上的结论有:dy\/dx=d[f(e^x)]\/dx*e^g(x)+f(e^x)*d[e^g(x)]\/dx= ...
设函数f(x)可导,且y=f(x2),则 dy\/dx=?
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)
设f(x)为可导函数,y=sin{f[sinf(x)]} dy\/dx=
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)]*cosf(x)*f'(x)
已知f(x)可导.y=f(sinx) 求dy\/dx ..
dy\/dx=[d(f(sinx))\/d(sinx)](dsinx\/dx)=(cosx)[d(f(sinx))\/d(sinx)]
设f(x),g(x)可导,y=f(e^x)e^g(x),求dy\/dx
(1)乘积的微分,y=f(x)g(x),则dy\/dx=f'(x)g(x)+f(x)g'(x)(2)复合函数的微分,链式法则,y=f(g(x)),dy\/dx=f'(g(x))g'(x)根据以上的结论有:dy\/dx=d[f(e^x)]\/dx*e^g(x)+f(e^x)*d[e^g(x)]\/dx = f'(e^x)e^x*e^g(x)+f(e^x)*e^g(x)*g'(...
设f(x)可导,且y=f(x^2),则dy\/dx=
y = f(x^2)dy\/dx = 2xf'(x^2)
设F(x)可导,y=f(x^2),则dy\/dx=?
根据复合函数求导法则 dy\/dx = [f(x^2)]' =f'(x^2) *(x^2)' = 2xf'(x)
设f(x)为可导函数,求dy\/dx (1)y=f(tanx) (2)y=f(x^2)+lnf(x)
1) y'=f'(tanx)* (tanx)'=f'(tanx) *(secx)^2 2) y'=f'(x^2)*2x+f'(x)\/f(x)