求分段函数的导数设 g(x)/x =0 f(x)={ 0 x=0 g(0)=g'(0)=0 g"(0)=2求 f‘(0)
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/07/16 04:31:31
求分段函数的导数
设 g(x)/x =0
f(x)={ 0 x=0 g(0)=g'(0)=0 g"(0)=2
求 f‘(0)
设 g(x)/x =0
f(x)={ 0 x=0 g(0)=g'(0)=0 g"(0)=2
求 f‘(0)
![求分段函数的导数设 g(x)/x =0 f(x)={ 0 x=0 g(0)=g'(0)=0 g](/uploads/image/z/3985961-41-1.jpg?t=%E6%B1%82%E5%88%86%E6%AE%B5%E5%87%BD%E6%95%B0%E7%9A%84%E5%AF%BC%E6%95%B0%E8%AE%BE+g%28x%29%2Fx+%3D0+f%28x%29%3D%7B+0+x%3D0+g%280%29%3Dg%27%280%29%3D0+g%22%280%29%3D2%E6%B1%82+f%E2%80%98%EF%BC%880%EF%BC%89)
f(x)=g(x)/x,x!=0;f(x)=o,x=0
g(x)在x=0的Taylor展开式为个g(x)=g(0)+g'(0)x+g"(0)x^2/2!+O(x^2)
f'(0)=lim(x->0)[(f(x)-f(0))/(x-0)]
=lim(x->0)f(x)/x
=lim(x->0)g(x)/x^2
=lim(x->0)[g(0)+g'(0)x+g"(0)x^2/2!+O(x^2)] /x^2
代入 g(0)=g'(0)=0 ,g"(0)=2,
=1
g(x)在x=0的Taylor展开式为个g(x)=g(0)+g'(0)x+g"(0)x^2/2!+O(x^2)
f'(0)=lim(x->0)[(f(x)-f(0))/(x-0)]
=lim(x->0)f(x)/x
=lim(x->0)g(x)/x^2
=lim(x->0)[g(0)+g'(0)x+g"(0)x^2/2!+O(x^2)] /x^2
代入 g(0)=g'(0)=0 ,g"(0)=2,
=1
求分段函数的导数设 g(x)/x =0 f(x)={ 0 x=0 g(0)=g'(0)=0 g"(0)=2求 f‘(0)
已知函数f(x)=lnx,g(x)=a/x(a>0),设F(x)=f(x)+g(x) 求F(x)的单调区间
已知函数f(x)的定义域是【0,3】,设g(x)=f(2x)-f(x+2).求g(x)的解析式和定义域
有关函数可导性的讨论设g(x)在x=0的某领域内二阶可导且g(0)=0,研究分段函数f(x)=g(x)/x,x≠0;g‘
帮忙解道高数导数的题设函数f(x),g(x)满足f'(x)=g(x),g'(x)=2e^x - f(x),且f(0)=0
设f(X)具有2阶连续导数,且f(a)=0,g(x)=f(x)/x-a,x不等于a,g(x)=f'(a),x=a,求g'
设函数f(X)定义在(0,+∞)上,f(1)=0,导数f'(x)=1/x,g(x)=f(x)+f'(x) .
已知对任意实数x,有f(-x)= - f(x),g(-x)= - g(-x),且x>0时,f(x)的导数>0,g(x)的
设g(x)在x=0点连续,求f(x)=g(x)sinx在x=0点的导数
函数f(x)=loga(x+2),g(x)=loga(2-x),h(x)=f(x)+g(x),求方程h(x)=0的解
设f(x)=xg(x),其中g(x)在x=0处连续,且g(0)=1,试用导数定义求f'(0).
设函数f(x)=tx+(1-x)/t(t>0),g(t)为f(x)在[0,1]上的最小值,求函数g(x)的最大值