數學問題 differentiation

[匿名]

2009-11-12 3:27 am

1)find the rate of increase of y with respect to x when x=1/3丌rad, given that y=1/2sinx. what is the smallest positive value of x for which the rate of increase is zero?

2)find dy/dx when i)y^2=x^2-8x+16, ii)y^2/3=x

3)evaluate d^2/dx^2(1+7/x-2/x^2)

4a)differentiate the following expressions with respect to x simplifying the results where possible:

i) log e(x^2+5x-3), ii)cos^4 3x

b) for y=3x e^-2x find the value of x which makes d^2y/dx^2=0.

回答 (2)

用戶9112

2009-11-12 3:42 am

✔ 最佳答案

(1) y = (1/2)sin x
dy/dx = (1/2)cos x
when x = π/3, dy/dx = (1/2)cos(π/3) = 1/4
when dy/dx = 0, cos x = 0 => x = π/2
(2) (i) y^2 = x^2 - 8x + 16
y^2 = (x - 4)^2
y^2 - (x - 4)^2 = 0
(y + x - 4)(y - x + 4) = 0
y = 4 - x or y = x - 4
dy/dx = -1 or +1
(ii) y^(2/3) = x
y = x^(3/2)
dy/dx = (3/2)x^(1/2)
(3) d/dx [1 + 7/x - 2/x^2]
= -7/x^2 + 4/x^3
d/dx [-7/x^2 + 4/x^3]
= 14/x^3 - 12/x^4
(4ai) y = ln(x^2 + 5x - 3)
dy/dx = d[ln(x^2 + 5x - 3)] / d[x^2 + 5x - 3] * d[x^2 + 5x - 3]/dx
= (2x + 5)/(x^2 + 5x - 3)
(ii) y = cos^4(3x)
dy/dx = d[cos^4(3x)] / d[cos(3x)] * d[cos(3x)] / d[3x] * d[3x]/dx
= [4cos^3(3x)][-sin(3x)][3]
= -12sin(3x)cos^3(3x)
(b) y = 3xe^(-2x)
dy/dx = 3e^(-2x) - 6xe^(-2x)
d^2y/dx^2 = -6e^(-2x) - 6e^(-2x) + 12xe^(-2x)
= 12e^(-2x)(x - 1) = 0
=> x = 1

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用戶7880

2009-11-18 4:15 am

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