[中學] 高中的數學

[mm0002uk (mm00)]

sin^-1 (2x) + sin^-1 (y) = π/2

求dy/dx, when x = 1/4

sin^-1 (2x)可以變成

dy/dx = 1/(√(1-4x^2))

sin^-1 (y)則要怎麼辦?
一樣是用→ 1/(√(1-y^2))(dy/dx)
這樣的話 y要怎麼樣消掉呢？

感謝幫忙喔!!!!

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推 kai25       :沒錯阿  然後用第一條式子代 x=1/4 就可以算出y
→ kai25       :asin(1/2) + asin(y) = pi/2 -> y = sin(pi/3)
→ b99401121   :sin^-1(y)=pi/2-sin^-1(2x)
→ b99401121   :y=sin(pi/2-sin^-1(2x))
→ b99401121   :y=sin(pi/2)cos(sin^-1(2x))-cos(pi/2)sin(sin^-1(2x
→ b99401121   :y=0+(1)*(2x)
→ b99401121   :dy/dx = 2
→ b99401121   :不好意思，dy/dx = -2
→ b99401121   :oh no
→ b99401121   :算錯了
→ b99401121   :y = cos(sin^-1(2x)) + 0
→ b99401121   :y = (1 - (sin(sin^-1(2x))^2)^1/2
→ b99401121   :y = (1-4x^2)^1/2
→ b99401121   :etc.