Re: [考古] 再幫個忙
※ 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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