How do we take the derivative here? y = square root(x^2 + 4x) − x?
回答 (3)
2016-05-01 12:34 pm
2016-05-01 1:24 pm
y = [ x² + 4x ]^(1/2) - x
dy/dx = (1/2) [ x² + 4x ]^(-1/2) [ x² + 4x ] - 1
dy/dx = [ x² + 4x ] / [ 2 (x² + 4x)^(1/2) ] - 1
dy/dx = (1/2) [ x² + 4x ]^(-1/2) [ x² + 4x ] - 1
dy/dx = [ x² + 4x ] / [ 2 (x² + 4x)^(1/2) ] - 1
2016-05-01 12:39 pm
y = sqrt(x^2+4x) - x
y = (x^2+4x)^(1/2) - x
y' = (1/2) (x^2+4x)^(-1/2) d/dx (x^2+4x) - 1
y' = (1/2)(x^2+4x)^(-1/2) (2x+4) - 1
y' = (2x+4) / (2(x^2+4x)^(1/2)) - 1
y' = (2x+4) /(2sqrt(x^2+4x)) - 1
y' = (x+2) / sqrt(x^2+4x) - 1
y = (x^2+4x)^(1/2) - x
y' = (1/2) (x^2+4x)^(-1/2) d/dx (x^2+4x) - 1
y' = (1/2)(x^2+4x)^(-1/2) (2x+4) - 1
y' = (2x+4) / (2(x^2+4x)^(1/2)) - 1
y' = (2x+4) /(2sqrt(x^2+4x)) - 1
y' = (x+2) / sqrt(x^2+4x) - 1
收錄日期: 2021-04-23 20:58:44
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160501042919AA5v4F7