设y=y(x)y = y(x)y=y(x)是由方程ln(x+y)+5=y\ln(x + y)+ 5 = yln(x+y)+5=y确定的隐函数,则y′=1x+y−1y'=\dfrac{1}{x + y - 1}y′=x+y−11.( )
已知函数y=f(x)y = f(x)y=f(x)二阶可导,且f′′(x0)=0f''(x_0) = 0f′′(x0)=0,则(x0,f(x0))(x_0,f(x_0))(x0,f(x0))是曲线y=f(x)y = f(x)y=f(x)的拐点.( )
已知f′(x)=x2f'(x)=x^2f′(x)=x2,则f(x)=x3f(x)=x^3f(x)=x3.( )
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