標題:

ce程度微分,急(40分)

發問:

y= tan^2 x 用y表示d^2y/dx^2 最好不要skip步驟,本人比較頓 記住用y表示,不是x thx

最佳解答:

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dy/dx = 2 tanx sec^2 x d^2y/dx^2 = 2tanx (2secx tanx secx) + 2sec^2 x (sec^2 x) = 4tan^2 x sec^2 x + 2sec^4 x = 4tan^2 x (1+ tan^2 x ) + 2(1+ tan^2 x)^2 =4y(1+ y) + 2(1 +y)^2 =4y + 4 y^2 + 2 (1+ 2y +y^2) =4y + 4 y^2 + 2+ 4y +2 y^2 =6 y^2 + 8y+ 2

其他解答:

dy/dx = (2tanx) (sec^2 x) = 2tanx sec^2 x d^2y/dx^2 = 2 [(tanx)(2secx)(secx tanx) + (sec^2 x)(sec^2 x)] = 2 (tan^2 x sec^2 x + sec^4 x) = 2sec^2 x (tan^2 x + sec^2 x) = 2sec^2 x (y + sec^2 x) .........Sub. y= tan^2 x into the equation...... d^2y/dx^2 / 2sec^2 x = y + sec^2 x y = d^2y/dx^2 / 2sec^2 x - sec^2 x|||||dy/dx = 2 tanx sec^2 x d^2y/dx^2 = 2tanx (2secx tanx secx) + 2sec^2 x (sec^2 x) = 4tan^2 x sec^2 x + 2sec^4 x = 4tan^2 x (1+ tan^2 x ) + 2(1+ tan^2 x)^2 =4y(1+ y) + 2(1 +y)^2 =4y + 4 y^2 + 2 (1+ 2y +y^2) =4y + 4 y^2 + 2+ 4y +2 y^2 =6 y^2 + 8y+ 2
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