Please help with the following differentiation. If y=sec^2(x) show that dy/dx =2y (sqrt (y-1))?
回答 (3)
2017-09-03 12:48 am
✔ 最佳答案
Let u = sec xy = u^2
dy/dx = 2u du = 2secx * secxtanx = 2sec^2x*tanx = 2y*tan(x)
Since tan^2x + 1 = sec^2x, tanx = sqrt(sec^2(x) - 1) = sqrt(y-1), thus dy/dx = 2y*sqrt(y-1)
2017-09-02 11:33 pm
y = sec(x)^2
y^(1/2) = sec(x)
(1/2) * y^(-1/2) * dy = sec(x) * tan(x) * dx
dy/dx = 2 * sec(x) * tan(x) / y^(-1/2)
dy/dx = 2 * y^(1/2) * sec(x) * tan(x)
dy/dx = 2 * y^(1/2) * y^(1/2) * sqrt(sec(x)^2 - 1)
dy/dx = 2 * y * sqrt(y - 1)
y^(1/2) = sec(x)
(1/2) * y^(-1/2) * dy = sec(x) * tan(x) * dx
dy/dx = 2 * sec(x) * tan(x) / y^(-1/2)
dy/dx = 2 * y^(1/2) * sec(x) * tan(x)
dy/dx = 2 * y^(1/2) * y^(1/2) * sqrt(sec(x)^2 - 1)
dy/dx = 2 * y * sqrt(y - 1)
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