函数的证明题.
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/06/23 12:42:20
函数的证明题.
![](http://img.wesiedu.com/upload/e/46/e46c49a4fa62935fcf2008648ba6bad8.jpg)
![](http://img.wesiedu.com/upload/e/46/e46c49a4fa62935fcf2008648ba6bad8.jpg)
![函数的证明题.](/uploads/image/z/3313854-54-4.jpg?t=%E5%87%BD%E6%95%B0%E7%9A%84%E8%AF%81%E6%98%8E%E9%A2%98.)
用求导法比较方便
设f(x)=arctanx-1/2arcos(2x/1+x^2)
求得:f'(x) =1/(1+x^2)-1/2*(-1)*1/√1-(2x/(1+x^2))^2*(2x/(1+x^2))'=0
那么f(x)=C
f(1) = π/4 - (1/2) arccos[2/(1+1)]=π/4 - (1/2)arccos1= π/4 - 0= π/4.
那么f(x) = π/4.
即知命题成立.
再问: f'(x)的算法能写详细点么?? 看不懂。。。
再答: f'(x)=(arctanx-1/2arcos(2x/1+x^2))' =(arctanx)'-1/2(arccos(2x/(1+x^2))' =1/(1+x^2)-1/2*(-1)*1/√1-(2x/(1+x^2))^2*(2x/(1+x^2))' =1/(1+x^2)-1/2*(-1)*(x^2-1)/(x^2+1)*2*(1-x^2)/(x^2+1)^2 =1/(1+x^2)-1/(1+x^2) =0 够详细了吧
设f(x)=arctanx-1/2arcos(2x/1+x^2)
求得:f'(x) =1/(1+x^2)-1/2*(-1)*1/√1-(2x/(1+x^2))^2*(2x/(1+x^2))'=0
那么f(x)=C
f(1) = π/4 - (1/2) arccos[2/(1+1)]=π/4 - (1/2)arccos1= π/4 - 0= π/4.
那么f(x) = π/4.
即知命题成立.
再问: f'(x)的算法能写详细点么?? 看不懂。。。
再答: f'(x)=(arctanx-1/2arcos(2x/1+x^2))' =(arctanx)'-1/2(arccos(2x/(1+x^2))' =1/(1+x^2)-1/2*(-1)*1/√1-(2x/(1+x^2))^2*(2x/(1+x^2))' =1/(1+x^2)-1/2*(-1)*(x^2-1)/(x^2+1)*2*(1-x^2)/(x^2+1)^2 =1/(1+x^2)-1/(1+x^2) =0 够详细了吧