一道导数的题目f(x)在(0,+∞)有定义,对于任意x∈(0,+∞),y∈(0,+∞),有f(xy)=f(x)+f(y)+(x-1)(y-1),又f '(1)=a≠1.证明对任意x∈(0,+∞),f '(x)存在并求之
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/14 16:06:25
![一道导数的题目f(x)在(0,+∞)有定义,对于任意x∈(0,+∞),y∈(0,+∞),有f(xy)=f(x)+f(y)+(x-1)(y-1),又f '(1)=a≠1.证明对任意x∈(0,+∞),f '(x)存在并求之](/uploads/image/z/7215249-57-9.jpg?t=%E4%B8%80%E9%81%93%E5%AF%BC%E6%95%B0%E7%9A%84%E9%A2%98%E7%9B%AEf%28x%29%E5%9C%A8%EF%BC%880%2C%2B%E2%88%9E%EF%BC%89%E6%9C%89%E5%AE%9A%E4%B9%89%2C%E5%AF%B9%E4%BA%8E%E4%BB%BB%E6%84%8Fx%E2%88%88%280%2C%2B%E2%88%9E%29%2Cy%E2%88%88%280%2C%2B%E2%88%9E%29%2C%E6%9C%89f%28xy%29%3Df%28x%29%2Bf%28y%29%2B%28x-1%29%28y-1%29%2C%E5%8F%88f+%27%281%29%3Da%E2%89%A01.%E8%AF%81%E6%98%8E%E5%AF%B9%E4%BB%BB%E6%84%8Fx%E2%88%88%280%2C%2B%E2%88%9E%29%2Cf+%27%28x%29%E5%AD%98%E5%9C%A8%E5%B9%B6%E6%B1%82%E4%B9%8B)
一道导数的题目f(x)在(0,+∞)有定义,对于任意x∈(0,+∞),y∈(0,+∞),有f(xy)=f(x)+f(y)+(x-1)(y-1),又f '(1)=a≠1.证明对任意x∈(0,+∞),f '(x)存在并求之
一道导数的题目
f(x)在(0,+∞)有定义,对于任意x∈(0,+∞),y∈(0,+∞),有f(xy)=f(x)+f(y)+(x-1)(y-1),又f '(1)=a≠1.
证明对任意x∈(0,+∞),f '(x)存在并求之
一道导数的题目f(x)在(0,+∞)有定义,对于任意x∈(0,+∞),y∈(0,+∞),有f(xy)=f(x)+f(y)+(x-1)(y-1),又f '(1)=a≠1.证明对任意x∈(0,+∞),f '(x)存在并求之
1+(a-1)/x
这个.问问别人吧! 题目? f(x)=ax^3+2ln(1-x)(x<1) f'(x)=3ax^2-2/(1-x)>=0 3a>=2/[x^2(1-x)] 设h(x)=x^2(1-x
神马神马,好晕
因f(xy)=f(x)+f(y)+(x-1)(y-1)
可取y=(x+Δx)/x =1+Δx/x
f[x(1+Δx/x)]=f(x)+f(1+Δx/x)+(x-1)(1+Δx/x-1)
即 f[x+Δx] - f(x) = f(1+Δx/x)+Δx (x-1) /x
{f[x+Δx] - f(x) } /Δx = f(1+Δx/x)/Δx + (x-1) /x
全部展开
因f(xy)=f(x)+f(y)+(x-1)(y-1)
可取y=(x+Δx)/x =1+Δx/x
f[x(1+Δx/x)]=f(x)+f(1+Δx/x)+(x-1)(1+Δx/x-1)
即 f[x+Δx] - f(x) = f(1+Δx/x)+Δx (x-1) /x
{f[x+Δx] - f(x) } /Δx = f(1+Δx/x)/Δx + (x-1) /x
所以 Δx→0 lim {f[x+Δx] - f(x) } /Δx
= lim f(1+Δx/x)/Δx + (x-1) /x
= (x-1) /x +(1/x)* lim f(1+Δx/x) /(Δx/x)
又f '(1)=a≠1=lim f(1+Δx/x) /(Δx/x)
所以
f '(x)=lim {f[x+Δx] - f(x) } /Δx
=(x-1) /x +(1/x)* f '(1)
=1+(a-1)/x
收起