设f(x)可导y=f(cos2x) dy=-搜答案-拍题作业网

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

y'=f'(sin²x)*(sin²x)'+f'(cos²x)*(cos²x)'=f'(sin²x)*(2sinxcos)+f'(cos²x

见图,复合函数求导.

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

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

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

y'=f'(e^(-2x)+cosx)(e^(-2x)'+cos'x)=f'(e^(-2x)+cosx)(-2e^(-2x)-sinx)

d/dx(f(sin^2(x))+sin(f(x)^2)) = sin(2 x) f'(sin^2(x))+2 f(x) f'

Z'x=-yf'(y/x)y/x^2xZ'=-y^2f'(y/x)/xZ'y=xf'(y/x)1/xyZ'y=yf'(y/x)xZ'x+yZ'y=-y^2f'(y/x)/x+yf'(y/x)=y(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)]

A

eZ/eX=2x*[ef(x*x-y*y)/ex],eZ/eY=-2x*[ef(x*x-y*y)/ey],

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)

y=f(x-y)dy/dx=f'(x-y)*d(x-y)=f'(x-y)*(1-dy/dx)=f'(x-y)-f'(x-y)*dy/dx[1+f'(x-y)]dy/dx=f'(x-y)dy/dx=f'

若看不清楚,可点击放大.

∂z/∂x=-((∂f/∂x)*y*2x)/f^2∂z/∂y=1/f+2y2*(∂f/∂y)/f^21/

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

y'=f'(sin(2x))*(sin(2x))'+(sin(f(2x)))'*f'(2x)=f'(sin(2x))*2*cos(2x)+cos(f(2