f(x)=ln(1除以x^2 4x 5)展开成(x-2)的幂级数
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![f(x)=ln(1除以x^2 4x 5)展开成(x-2)的幂级数](/uploads/image/f/575218-10-8.jpg?t=f%28x%29%3Dln%281%E9%99%A4%E4%BB%A5x%5E2+4x+5%29%E5%B1%95%E5%BC%80%E6%88%90%28x-2%29%E7%9A%84%E5%B9%82%E7%BA%A7%E6%95%B0)
F(X)的导数=2X/(1+X^2)
楼主想求什么?f(x)的极限吗?如果是求极限,那么可得f(x)在x趋于无穷大的时候极限为0.
再答:对数函数的真数恒正x+1>0所以定义域为x>-1再问:已知i是虚数单位,若(m+i)^2=3-4i,则实数m的值为再答:再答:别只看答案,看看过程,那里不懂问问,超个答案下次还不会哦再问:能再问
复合函数的求导令x²-1=tf(x)=Intf'(x)=Int'*t'=1/(x²-1)*2x=2x/(x²-1)
lim(x→0)f(x)/x=f'(0)=1再问:我没看明白哎求解。。再答:lim(x→0)f(x)/x=lim(x→0)[f(x)-f(0)]/(x-0)=f'(0)=1
f'(x)=1/(x+1)+a>=2xa>=2x+1/(x+1)g(x)=2x+1/(x+1)g'(x)=2-1/(x+1)²1
①f(x)=ln(x+1)定义域(-1,+∞)f(0)=0在(0,+∞)存在一点ε,0<ε<1/xf(1/x)-f(0)=f'(ε)(1/x-0)f'(x)=1/(x+1)∵0<ε<1/x∴1/(1/
解题思路:利用导数求最值解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/readq
f(x)=ln√(x²+1)f'(x)=[1/√(x²+1)]*(√(x²+1))'=[1/√(x²+1)]*[1/2√(x²+1)]*(x²
令g(x)=f(x)-ax-b=ln(x+1)-(a+2)x+2-b≤0;再令g'(x)=[1/(x+1)]-(a+2)=0,求得g(x)的驻点(当a>-2时是极大值点):x0=-(a+1)/(a+2
1.f'(x)=e^x-1/(x+1),f'(0)=0,f''(x)=e^x+1/(x+1)^2>0,f'(x)为(-1,+∞)上的增函数,所以x>0时,f'(x)>f'(0)=0,f(x)在(0,+
f(x)=ln(x+√1+x^2)f'(x)=1/(x+√(1+x^2)*(x+√1+x^2)'=1/(x+√(1+x^2)*(1+(√1+x^2)'=1/(x+√(1+x^2)*(1+1/2*√(x
f(x)=ln(x+1),则:f'(x)=[1/(x+1)]×(x+1)'.=[1/(x+1)]×1.=1/(x+1)
楼主这么晚还没休息啊我想请问一下楼主的f(x)=ln(1+x)/x//ln(1+x)是从网上看到的?还是从书本上看到的?而且,我认为,楼主f(x)=ln(1+x)/x//ln(1+x)打多了一个除号,
(uv)'=u'v+v'uf'(x)=ln(x+1)+1
f(x)=ln(x-1)+2a/x,f'(x)=1/(x-1)-2a/x^2=[x^2-2a(x-1)]/[(x-1)x^2]=(x^2-2ax+2a)/[(x-1)x^2](x>1),a0,f(x)
f(x)=ln(1+x)-ln(1-x)=ln2f(x)=ln(1+x)/(1-x)=ln2(1+x)/(1-x)=2(1+x)=2(1-x)1+x=2-2x3x=1x=1/3
f'(x)=1-1/(x+1)=x/(x+1)再问:你没看到图片吗再答:没有图片啊