Question

f(x)=ax/(x^2+b)怎样求导？需要每一步的详细过程，谢谢！急~~~

Answer 1

f'(x)=[a(x^2+b)-2ax^2]/(x^2+b)^2
即原分母平方，原分子求导乘以原分母，再减去原分母求导乘以原分子

Answer 2

先求倒数再求导，我觉得这样做不容易出错
令y=f(x)，则1/y=(x^2+b)/ax=x/a+b/ax
对两边求导，得
-y'/y^2=1/a-b/ax^2=(x^2-b)/ax^2
y'=y^2(b-x^2)/ax^2
将y=f(x)=ax/(x^2+b)代入并化简，可得
y'=a(b-x^2)/(x^2+b)^2

Answer 3

f(x)=ax/(x^2+b)
f`(x)={a(x^2+b)-ax(2x)}/(x^2+b)^2
={ax^2+ab-2ax^2}/(x^2+b)^2
={ab-ax^2}/(x^2+b)^2

Answer 4

f=7a=5

Answer 5

[ax/(x^2+b)]'= a[x/(x^2+b)]' =a[x'*(x^2+b)-(x^2+b)'*x]/(x^2+b)^2=a[(x^2+b)-2x*x]/(x^2+b)^2