please help please. find the derivative 1)dy/dx if y(x)= square root x^2+9?
回答 (4)
2016-06-21 3:42 pm
1)
y = √(x² + 9)
dy/dx
= d[√(x² + 9)]/dx
= {d[√(x² + 9)]/d(x² + 9)} • [d(x² + 9)/dx]
= {1/[2√(x² + 9)]} • (2x)
= x/√(x² + 9)
y = √(x² + 9)
dy/dx
= d[√(x² + 9)]/dx
= {d[√(x² + 9)]/d(x² + 9)} • [d(x² + 9)/dx]
= {1/[2√(x² + 9)]} • (2x)
= x/√(x² + 9)
2016-06-21 5:19 pm
y = [ x² + 9 ]^(1/2)
dy/dx = (1/2) [ x² + 9 ]^(-1/2) (2x)
dy/dx = x / [ x² + 9 ]^(1/2)
dy/dx = (1/2) [ x² + 9 ]^(-1/2) (2x)
dy/dx = x / [ x² + 9 ]^(1/2)
2016-06-21 4:17 pm
y= sqrt(x^2+9)
dy/dx = 1/ (2sqrt(x^2+9) ) d/dx (x^2+9)
= 2x / (2 sqrt(x^2+9))
= x / sqrt(x^2+9)
dy/dx = 1/ (2sqrt(x^2+9) ) d/dx (x^2+9)
= 2x / (2 sqrt(x^2+9))
= x / sqrt(x^2+9)
2016-06-21 3:40 pm
y = (x^2 + 9)^(1/2)
y^2 = x^2 + 9
2y * dy = 2x * dx
dy/dx = x/y
dy/dx = x / sqrt(x^2 + 9)
y^2 = x^2 + 9
2y * dy = 2x * dx
dy/dx = x/y
dy/dx = x / sqrt(x^2 + 9)
收錄日期: 2021-04-18 15:10:16
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160621073728AA24Rrs