我想問關於4條Differentiation問題點計

lim ( 8x^2+10x+12)/(2x^2+1)

=lim (8+10/x+12/x^2)/(2+1/x^2)

=8/2

=4//

y=(a+8)/a

dy/da= [a-(a+8)]/a^2 = -8/a^2

da/dx=2(4x^2+10) (8x)

dy/dx

=dy/da x da/dx (by chain rule)

= -8/a^2 x 2(4x^2+10) (8x)

=-128/(4x^2+10)^3//

y=14sin6x+6cos14x

dy/dx=14(6)cos6x+6(14)(-sin14x)=84(cos6x-sin14x)//

y=6x^2-34x+90

dy/dx=6(2x)-34=12x-34//

y=(x^2+3x+2)(x+1) =(x+1)(x+2)/(x+1)=x+2

dy/dx=1//


2009-06-01 22:24:21 補充：
y=(x^2+3x+2)(x+1)

y=(x+2)(x+1)^2

dy/dx=2(x+1)(x+2)+(x+1)=(x+1)(2x+5)=2x^2+7x+5//

Q1.
If tends to zero, the answer is (0 + 0 + 12)/(0 + 1) = 12/1 = 12.
If tends to infinity, the expression = [x^2(8 + 10/x + 12/x^2)]/x^2(2 + 1/x^2) = (8 + 10/x + 12/x^2)/(2 + 1/x^2) = (8 + 0 + 0)/(2 + 0) = 8/2 = 4.
Q2.
y = (a + 8)/a = 1 + 8/a, so dy/da = -8/a^2.
a = (4x^2 + 10)^2, so da/dx = 2(4x^2 + 10)(8x) = 16x(4x^2 + 10).
dy/dx = (dy/da)(da/dx) = (-8/a^2)(16x)(4x^2 + 10).
Since y = (a + 8)/a
ay - a = 8
a(y - 1) = 8
a = 8/(y -1)
so dy/dx = -2x(y - 1)^2(4x^2 + 10).
Q3.
y = 14sin6x + 6cos14x
dy/dx = (14cos6x)(6) - (6sin14x)(14)
= 84(cos 6x - sin 14x).
Q4.
i)dy/dx = 12x - 34.
ii) dy/dx = (x + 1)(2x + 3) + (x^2 + 3x + 2)
= 2x^2 + 5x + 3 + x^2 + 3x + 2 = 3x^2 + 8x + 5.