设f(0)的二阶导数存在,且f(0)=0,g(x)=f(x)/x (x≠0时) g(x)=f(0)的导数(x=0时),则g(0)的导数为如题设f(0)的二阶导数存在,且f(0)=0,g(x)=f(x)/x,(x≠0时) g(x)=f(0)的导数,(x=0时),则g(0)的导数为多
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![设f(0)的二阶导数存在,且f(0)=0,g(x)=f(x)/x (x≠0时) g(x)=f(0)的导数(x=0时),则g(0)的导数为如题设f(0)的二阶导数存在,且f(0)=0,g(x)=f(x)/x,(x≠0时) g(x)=f(0)的导数,(x=0时),则g(0)的导数为多](/uploads/image/z/9291853-37-3.jpg?t=%E8%AE%BEf%EF%BC%880%EF%BC%89%E7%9A%84%E4%BA%8C%E9%98%B6%E5%AF%BC%E6%95%B0%E5%AD%98%E5%9C%A8%2C%E4%B8%94f%280%29%3D0%2Cg%28x%29%3Df%28x%29%2Fx+%28x%E2%89%A00%E6%97%B6%29+g%28x%29%3Df%EF%BC%880%EF%BC%89%E7%9A%84%E5%AF%BC%E6%95%B0%EF%BC%88x%3D0%E6%97%B6%EF%BC%89%2C%E5%88%99g%280%29%E7%9A%84%E5%AF%BC%E6%95%B0%E4%B8%BA%E5%A6%82%E9%A2%98%E8%AE%BEf%EF%BC%880%EF%BC%89%E7%9A%84%E4%BA%8C%E9%98%B6%E5%AF%BC%E6%95%B0%E5%AD%98%E5%9C%A8%2C%E4%B8%94f%280%29%3D0%2Cg%28x%29%3Df%28x%29%2Fx%2C%28x%E2%89%A00%E6%97%B6%29+g%28x%29%3Df%EF%BC%880%EF%BC%89%E7%9A%84%E5%AF%BC%E6%95%B0%2C%EF%BC%88x%3D0%E6%97%B6%EF%BC%89%2C%E5%88%99g%280%29%E7%9A%84%E5%AF%BC%E6%95%B0%E4%B8%BA%E5%A4%9A)
设f(0)的二阶导数存在,且f(0)=0,g(x)=f(x)/x (x≠0时) g(x)=f(0)的导数(x=0时),则g(0)的导数为如题设f(0)的二阶导数存在,且f(0)=0,g(x)=f(x)/x,(x≠0时) g(x)=f(0)的导数,(x=0时),则g(0)的导数为多
设f(0)的二阶导数存在,且f(0)=0,g(x)=f(x)/x (x≠0时) g(x)=f(0)的导数(x=0时),则g(0)的导数为
如题设f(0)的二阶导数存在,且f(0)=0,g(x)=f(x)/x,(x≠0时) g(x)=f(0)的导数,(x=0时),则g(0)的导数为多少
设f(0)的二阶导数存在,且f(0)=0,g(x)=f(x)/x (x≠0时) g(x)=f(0)的导数(x=0时),则g(0)的导数为如题设f(0)的二阶导数存在,且f(0)=0,g(x)=f(x)/x,(x≠0时) g(x)=f(0)的导数,(x=0时),则g(0)的导数为多
由导数的定义有
g'(0)=lim(x-->0)[g(x)-g(0)]/(x-0)=lim(x-->0)[g(x)-g(0)]/x=lim(x-->0)[g(x)-f'(0)]/x
又因为当x不等于0时,有g(x)=f(x)/x,所以
g'(0)=lim(x-->0)[f(x)/x-f'(0)]/x=lim(x-->0)[f(x)-x*f'(0)]/x^2
因为该式的极限为0/0型,所以由罗必达法则(即所求极限等于分母的导数除以分子的导数)有
g'(0)=lim(x-->0)[f'(x)-f'(0)]/2x,
又因为该式的极限是0/0型,所以再次应用罗必达法则有
g'(0)=lim(x-->0)f''(x)/2=f''(0)/2
由于f'(0)=lim [f(x)-f(0)]/(x-0)=lim [f(x)-0]/x=lim f(x)/x
x->0 x->0 x->0
故
g'(0)=lim [g(x)-g(0)]/(x-0)=lim [f(x)/x-f'(0]/x("0/0"型,用洛必达...
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由于f'(0)=lim [f(x)-f(0)]/(x-0)=lim [f(x)-0]/x=lim f(x)/x
x->0 x->0 x->0
故
g'(0)=lim [g(x)-g(0)]/(x-0)=lim [f(x)/x-f'(0]/x("0/0"型,用洛必达法则)=lim [f'(x)x-f(x)]/x^2
x->0 x->0 x->0
("0/0"型,用洛必达法则)=lim[f''(x)x+f'(x)-f'(x)]/2x=limf''(x)/2=f''(0)/2
x->0 x->0
收起
f''(0)/2