导数微分实问题

f'(x)=2e^(2x)f[e^(-2x)]+e^(2x)f'[e^(-2x)]*e^(-2x)*(-2)

=2e^(2x)f[e^(-2x)]-2f'[e^(-2x)]

2.因为x = y ^ 2+y，所以:

u=[(y^2+y)^2+(y^2+y)]^(3/2)

所以:

u^(2/3)=(y^2+y)^2+y^2+y

两边同时求导给出:

(2/3)u^(-1/3)du=2(y^2+y)(2ydy+dy)+2ydy+dy

(2/3)u^(-1/3)du=(2y^2+2y+1)(2y+1)dy

所以:

dy/du=(2y^2+2y+1)(2y+1)/(2/3)u^(-1/3)

3:x^y=y^x

同时取两边的对数得到:

ylnx=xlny

派生以获得:

lnxdy+ydx/x=lnydx+xdy/y

(lnx-x/y)dy=(lny-y/x)dx

因此

dy/dx=(lny-y/x)/(lnx-x/y)

4:y=f(反正切)

y'=f'(arctanx)(arctanx)'

=f'(arctanx)/(1+x^2)

y''=[f''(arctanx)(1+x^2)-f'(arctanx)(1+x^2)']/(x^2+1)^2

=[f''(arctanx)(1+x^2)-2xf'(arctanx)]/(x^2+1)^2

5:y=e^[sin(1/x)]

y'=e^[sin(1/x)]*sin'(1/x)

=e^[sin(1/x)]*cos(1/x)*(-1/x^2)

=-e^[sin(1/x)]cos(1/x)/x^2

y''={2xe^[sin(1/x)cos(1/x)-[e^[sin(1/x)]cos(1/x)]'}/x^4

={2xe^[sin(1/x)cos(1/x)-[e^[sin(1/x)]*cos(1/x)*(-1/x^2)cos(1/x)+e^[sin(1/x)][-sin(1/x)](-1/x^2)]}/x^4

={2xe^[sin(1/x)cos(1/x)+(1/x^2)e^[sin(1/x)][cos^2(1/x)-sin(1/x)]}/x^4

所以:

d^2y={2xe^[sin(1/x)cos(1/x)+(1/x^2)e^[sin(1/x)][cos^2(1/x)-sin(1/x)]}dx/x^4.

6:uv=lnx*e^x

(uv)'=e^x/x+lnxe^x

(uv)''=(xe^x-e^x)/x^2+e^x/x+lnxe^x=(2x-1)e^x/x^2+lnxe^x

所以:

d^2(uv)=[(2x-1)e^x/x^2+lnxe^x]dx