1.对等式两边求导f'(x) = 2x + 2f'(1)当 x = 1 时,,,,,,,,f'(1) = 2 + 2f'(1)因此f'(1) = -2 ,,,f'(x) = 2x -4所以f'(0)=-4 2.f'(x)=g'(x)+2x因为曲线g(x)在点(1,g(1))处的切线方程为y=2x+1所以g'(1)=2所以f'(1)=g'(1)+2x1=4。即=f(x)在点(1,f(1))处切线的斜率为__4__ 3.用导数方法求解。设长A,则宽为S/A。周长y=2(A+S/A)y'=2-2S/A^2=0. 得到A=sqrt(S)时,y最小。 且ymin=2*2sqrt(S) 4.设该点
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