※ 引述《Aprilgreen (最後一年囉↗☆)》之銘言： : 若cos(x-y)=ysinx , dy/dx ? : 令 F(x,y) = cos(x-y) - ysinx dy Fx -sin(x-y) - ycosx sin(x-y) + ycosx 則 ---- = - ---- = - ------------------- = ------------------ dx Fy sin(x-y) - sinx sin(x-y) - sinx 若lim((√x^2-2x+5)-ax-b)=0.則(a,b)=? x→-∞ 此題等於在求√x^2-2x+5的其中一條漸近線 √x^2-2x+5 所以 a = lim ----------- x→-∞ x |x|√1-(2/x)+(5/(x^2)) = lim ----------------------- x→-∞ x -x√1-(2/x)+(5/(x^2)) = lim ----------------------- (因為x→-∞ 所以x < 0 => |x|=-x >0) x→-∞ x = -1 b = lim ((√x^2-2x+5) - (-1)*x) x→-∞ = lim ((√x^2-2x+5) + x) x→-∞ (√x^2-2x+5)^2 - x^2 = lim --------------------- x→-∞ (√x^2-2x+5) - x x^2-2x+5 - x^2 = lim ---------------------------- x→-∞ |x|√1-(2/x)+(5/(x^2)) - x -2x+5 = lim --------------------------- (x→-∞ => x < 0 => |x| = -x > 0) x→-∞ -x√1-(2/x)+(5/(x^2)) -x -2 + (5/x) -2 + 0 = lim --------------------------- = --------- = 1 x→-∞ -√1-(2/x)+(5/(x^2)) - 1 -1 - 1 所以 (a,b) = (-1,1) 若y=f(x)=x^7+3x+2,求(f的反函數)'(6)=? 令g(x)為f(x)的反函數 1 則 g(f(x)) = x => g'(f(x))f'(x) = 1 => g'(f(x)) = ------- f'(x) (f的反函數)'(6) = g'(6) 所以令 x^7+3x+2 = 6 => x = 1 f'(x) = 7x^6 + 3 => f'(1) = 7 + 3 = 10 1 1 因此(f的反函數)'(6) = g'(6) = ------- = ---- f'(1) 10 己知f'(x)=xe^x,f(0)=2 求f(x)=? f'(x)=xe^x => f(x) = ∫xe^x dx = xe^x - ∫e^x dx = xe^x - e^x + c 因為 f(0) = 2 所以 0 - 1 + c = 0 => c = 1 所以 f(x) = xe^x - e^x + 1 given the equation 2y=x^2+siny,find dy/dx ? 令 F(x,y) = 2y - x^2 - siny dy Fx -2x 2x 則 ---- = - ---- = - ---------- = ---------- dx Fy 2 - cosy 2 - cosy 讓f(x)=(tan的反函數)x,find the value of f'(1)? 1 f(x) = (tan的反函數)x => f'(x) = --------- 1 + x^2 1 1 所以 f'(1) = ------- = --- 1 + 1 2 --

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