[f(x)/g(x)] ' = 【f ' (x)·g(x) - f(x)·g ' (x)】/f^2(x).请证明一下.

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/17 11:52:50

[f(x)/g(x)] ' = 【f ' (x)·g(x) - f(x)·g ' (x)】/f^2(x).请证明一下.
[f(x)/g(x)] ' = 【f ' (x)·g(x) - f(x)·g ' (x)】/f^2(x).请证明一下.

[f(x)/g(x)] ' = 【f ' (x)·g(x) - f(x)·g ' (x)】/f^2(x).请证明一下.
[f(x+h)/g(x+h) - f(x)/g(x)] / h
=[f(x+h)g(x) - f(x)g(x+h)] / [hg(x)g(x+h)]
=[f(x+h)g(x) - f(x)g(x) + f(x)g(x) - f(x)g(x+h)] / [hg(x)g(x+h)]
=g(x)[f(x+h)-f(x)] / [hg(x)g(x+h)] - f(x)[g(x+h)-g(x)] / [hg(x)g(x+h)]
其中h趋于0得
=g(x)f'(x)/[g(x)]^2 - f(x)g'(x)/[g(x)]^2
=[g(x)f'(x)- f(x)g'(x)]/[g(x)]^2