1.已知,y=f[(3x-2)/(3x+2)],f'(x)=arctan(x^2),那么dy/dx(x=0)=?2.设函数y=y(x)由方程2^(xy)=x+y所确定,则dy(x=0)=?

Question

1.已知,y=f[(3x-2)/(3x+2)],f'(x)=arctan(x^2),那么dy/dx(x=0)=?
2.设函数y=y(x)由方程2^(xy)=x+y所确定,则dy(x=0)=?

Answer 1

1.已知,y=f[(3x-2)/(3x+2)],f'(x)=arctan(x^2),那么dy/dx(x=0)=?
dy/dx=f'[(3x-2)/(3x+2)]*{[(3x-2)/(3x+2)]}'
=f'[(3x-2)/(3x+2)]*12/(3x+2)^2
故dy/dx(x=0)=f'(-1)*12/4=3arctan1=3pi/4
2.设函数y=y(x)由方程2^(xy)=x+y所确定,则dy(x=0)=?
方程两边同时求自然对数：
xyln2=ln(x+y)
上方程两边对x求导：
yln2+xy'ln2=(1+y')/(x+y)
解得dy/dx=y'=[(xy+y^2)ln2-1]/[1-(x^2+xy)ln2]
故：dy={[(xy+y^2)ln2-1]/[1-(x^2+xy)ln2]}dx
当x=0时,由2^(xy)=x+y易知y(0)=0
所以：dy(x=0)=-dx